There is a serious problem in the proof as others before me mentioned. With a sum whose length goes to infinity, we are not allowed to take the limit term by term. Otherwise, you can prove that 1=0 by considering 1=n*(1/n)=(1/n)+(1/n)+(1/n)+... Taking the limit, we obtain 1=0+0+0....=0 In fact, the limit of the denominator should have been calculated using the squeeze theorem and noting that e^(-i)>=(1-i/n)^n>=e^(-i)*(1-i^2/n)
That was just a very simple example where things go wrong. In general, the theorem says that if {a_n} approaches the limit a and {b_n} approaches the limit b, then {a_n+b_n} approaces a+b. By induction you can extend this to any *fixed* number of summands. The theorem does not allow you to do it for a *variable* number of summands
Yinon Nahum thanks for the feedback. I was wondering the example you gave was in the indeterminate form of inf*zero. But the one that we had in this problem wasn't in any indeterminate form.
blackpenredpen the example is well defined. Taking the limit term by term, we obtain the sum of n zeros, which is equal to n*0=0. Regardless of the example, taking the limit term by term is allowed when one has a fixed number of summands. Doing so when the number of terms tends to infinity is not based on anything.
There's a slight logic error. When evaluating the bottom, you take the limit of each term as if n goes to infinity first without considering that the amount of terms is also depending on n. It's not clear that in general this kind of exchange of limit and sum works. To do this precisely, there is more care needed, though the final answer should stay the same.
Mike Montoro i just find an example with the sum from 1 to n of 1/n : always do 1 but if you goes to the limit in the term first you get 0... So your idea was really important !
Alright so here's what I've figured out: let a_k,n=(1-k/n)^n, A_k=e^-k. We need to show the following (some sort of uniform convergence): there exists an N such that for all k and all x>0, for all n
This is a very interesting limit problem. It’s hard to get an intuitive handle on it. One approach: Top is a geometric series, and is = (nⁿ⁺¹ - n)/(n-1) Bottom is a sum of the first n, n’th powers, which can be shown to be ∑₁ⁿ jⁿ = nⁿ⁺¹/(n+1) + ½ nⁿ + O(n) O(nⁿ⁻¹) + ... So there are n+1 terms, each of order nⁿ. This means that you get an infinite series whose terms are difficult to establish, multiplying nⁿ. And that makes the limit look like (n+1)(nⁿ⁺¹ - n) ----------------- (n-1)nⁿ⁺¹[1 + (n+1)/2n + O(1)] which in the limit, becomes 1/S, where S is that unknown series. Another approach to modeling the denominator is to approximate it with the integral from x=½ to n+½ of xⁿ. And that’s [(n+½)ⁿ⁺¹ - (½)ⁿ⁺¹]/(n+1) → (n+½)ⁿ⁺¹/(n+1) So now the limit looks like nⁿ⁺¹ ----- = [n/(n+½)]ⁿ⁺¹ = 1/[1 + 1/2n]ⁿ⁺¹ → 1/e^(½) = e^(-½) ≈ 0.60653... (n+½)ⁿ⁺¹ Of course, this is nowhere near rigorous; the interval of integration could be adjusted slightly, and the integral is after all, only an approximation. But I think this shows that the limit is finite and finitesimal; i.e., positive and finite. After watching: That was really cool! So my estimate was pretty close - that was unexpected. 1 - e⁻¹ = 0.63212... Thanks, bprp! Fred
I have seen the comment by @Mike Montoro (and there may be others that are similar), but there is not a slight error in logic in this proof but a HUGE one. The way that you take the bottom limit is not at all valid; you cannot take the limit term-wise when the number of terms depends on n. As a counterexample, consider the sum 1/(n^3) + 4/(n^3) + ... + (n^2)/(n^3). By your method, if we were able to send n to infinity for each term, then each term would go to zero and the sum of all would be zero. But this is not the limit of this sum! It can be re-written as (1/n)(1/n)^2 + (1/n)(2/n)^2 +... + (1/n)(n/n)^2. From here we can see that this is a Riemann sum of a function x^2 on a partition of n intervals of equal length between 0 and 1. So sending n to infinity would send the entire sum to the integral of x^2 from 0 to 1, which is 1/3; not 0.
Daniel Juncos Your argument is valid, but the problem is that he never took the limit term wise. Both the index of summation and the terms were taken to infinity simultaneously, which is what he was supposed to do.
Harder than I thought indeed. I thought this would be 0 as both series diverge and and imagine doing lahospital's rule over and over and the numerator should appraoch N! whereas the denominator would still have ln(n)^n(N^n) at least. I guess lahospital's rule can not work if the partial sum (the series itself)is changing. A great lesson indeed. Thank you.
I think the rule actually works correctly but the problem is taking the derivative properly is difficult. Because you're taking the derivative w.r.t. n and n is also the number of terms in the expression. On the top you can rewrite to a constant number of terms, but on the bottom it is not so straightforward.
It’s because you have assumed that it’s differentiable, when the above series isn’t, and lhopitals rule only works if you can find elementary derivatives
Great video, but I preferred it when you didn’t disclose the answer up front, but only at the end when you finished solving! Waiting for the answer is part of the fun :)
I have two requests for you. 1. Can you do a video about verifying whether the fourier series of a function converges uniformly? 2. Can you do something related to group theory? (e.g. finding whether given groups are isomorphic; finding composition series of a group; is a group simple? is a group solvable? What are the divisors of a given free abelian group? etc.. if these things appeal to you at all)
Not exactly a neat way, but I instead took that n^n as common on both the numerator and denominator and canceled them out. As the last terms become 1, I checked the value of previous term which is n-1. The numerator does fine as the previous term value ends up at 1/n, which just becomes 0 when equated. As for the denominator, the previous term (n-1) is (1-(1/n))^n, and the one before that will be (1-(2/n))^n. As I keep checking the previous terms, the value goes off to 0 so I just leave the denominator with three previous terms and the last term (1). Then I plug in the approximations for those values so I sort of end up with it 1/(1+0.1350+.36786+.04956) Which happened to converge to the original answer. Not neat one bit, but somehow converges to it.
When you take the limit in the bottom, it is NOT taking the limit of partial sums of a series - each term is changing with n, so it is not always true that you scan swap the order of limit and series sum. To do this, you need to prove (I believe) uniform convergence of the series.
Ashwin Singh The uniform convergence of the denominator is trivial though, since it is literally just Faulhaber's formula, which is well-known to have said properties. It's not like we have to prove Pythagoras' theorem everytime we use it.
Formally, the limit swap at 9:22 was possible due to "Tannery's theorem" - a result of the dominant convergence theorem. I guess that you dont want to talk about a little higher topics such as measure theory because most of your viewers probably aren't mathematicians, but ones that enjoy more fundamental math - and thats ok. But its not ok to make a video out of a problem you think your viewers will find interesting and provide a pretty detailed solution, even links to the theorems in the description which makes it disguised as a formal proof while you actually provides a half-wrong one. That's a misleading. You didnt have to make this a "heavy" video with measure theory - you just had to mention the theorem..
Hey, I'm not completely sure, but if Wikipedia's not lying about what Tannery's theorem is, then I assume it won't be helping here, because we can't uniformly dominate the coefficient sequences by any integrable sequence M_k.
Hey I got a question. Tannery's theorem is used for interchanging an infinite limit with an infinite summation. What trips me up is the summation in this video is not quite an infinite summation but rather the top of the "sigma" is also determined by n. You see this for a split moment at 7:59. As you can see the summand depends on n but also the number of terms. Is that a problem?
Agree, simply mention what theorem might come into play suffices. What irritates me more is that some comments say this is like "well-known", or saying "the result is still correct so no problem", just wanting to show off their knowledge? There are theorems for you to mention, not like a multiplication table that everyone knows.
@@hassanakhtar7874 you can just complete the infinite sum with infinite zero terms. In the notation of the Wikipedia page, define: a_k(n) = (1-k/n)ⁿ for 0 ≤ k ≤ n-1, and a_k(n) = 0 for k ≥ n. Then for n → +∞ lim a_k(n) = e⁻ᵏ, and |a_k(n)| = a_k(n) ≤ e⁻ᵏ = M_k, which satisfies the hypothesis of the theorem, and so the trick works.
To complete your proof you need to show that limite of sum ( k /n ) ^n converges to sum ( exp(-k)). First we can say that (k/n)^n = (1 - (n-k)/n) ^n . and then we can rearange the sum by doing the following change : k'=n-k So the sum (k/n)^n = sum (1- k'/n) Once this is done we can show that every term of the series can be majorated like that (1-k/n)^n
If you want, an other limit n-> infinity with the expression : 1*1 + 2*2 + 3*3 + 4*4 + .... + (n-1)*(n-1) + n*n (* : power to) ; this expression at any place in the fraction (up, down)
*Breathes in* I hope this is understandable in youtube writing format. Ok this theorem is called dominant convergence theorem and is a general measure theory result (it works for series, integrals, probabilities and more). Suppose there is a sequence a(n,m) such as a(n,m) converges to some a(m) when n alone approaches infinity. Now consider there is a sequence b(m) such as b(1)+b(2)+b(3)+... converges, but also b(m) >= |a(n,m)| for any n. Then you can totally swap the lim and the series sign. So you can get the full picture, a(n,m)= (1-m/n)^n Now we get to the point: how did BPRP know he could use this theorem? Because you can absolutely use a(m) as b(m) when a(n,m) is always positive (check) and ever increasing with respect to n: (1-m/n)^n
@@thedarkspeedninjashadittsux In the harmonic series, for instance. Or in a case such as the lim n -> ∞ [∑_{k=0}^{n} 1/n^p + 1/2^k * (∑_{i=0}^{n} 1/2^i)] where summing the limits of each term results in two, while the limit of the whole goes to infinity for Re(p) ≤ 1, and results in ζ(p) + 2 for Re(p) > 1.
with the top isnt it easier to factor out n^n and cancel everything on the inside of the brackets to get n^n(0+0+0+...0+1) since the 1/n^a's will approach 0 anyway, and you have 1 because n^n/n^n = 1, so the top becomes 1*n^n = n^n, and so with ur next step of dividing by n^n on the top and bottom you get the top being 1 with much less work
oh, that Fact didn't seem very intuitive, but for the ln of small numbers (close to 1) I do recall a limit: lim n to infinity of ln (1+a/n) = a/n, so indeed (1+a/n)^n = e^a So the bottom sum / n^n indeed becomes the sum of all these terms that approaches the sum 1+e^-1+... A bit more clear method of solving series would be nice, instead of using a rule like a magic wand. Ask what is needed to remove all the middle terms so the first and last term remain: 1. the series n+n^2+.. +n^n needs a factor 1-1/n or (n-1)/n to become n^n-1. bringing this factor to the other side and setting apart n^n, the sum equals n^n * n/(n-1)*(1-1/n^n), which will approach n^n * 1 in the limit. 2. the series 1+e^-1+e^-2+... needs a factor 1-1/e to become 1-e^-(n+1). ignoring this last term and bringing the factor to the other side the limit of the series becomes 1/(1-1/e).
Clever method to obtain Geometric sum series to infinity on the bottom - took me a while to figure out why you divide by n^n. :) Can you make a video on doing sum series for sinh^n ( or any hyperbolics)? Just for the maths :)
.what is done after 7:01 is wrong . Here is the counter example: integral of x on [0,1] =1/n^2+2/n^2+...+n/n^2 (You can find this by using Riemann sums). In fact al terms on the RHS of this equation foes to 0 as n goes to infinity. Hence, we have integral of x on[0,1] =0+0+0+...=0
Probably worth mentioning that the step at 7 minutes is dependent on n approaching infinity as the previous denominator is finite but the subsequent one(that gets summed up) is not?
If the numerator was just n^n, when you divide it by n^n, the result would still be 1. The other terms don't effect the result. I initially tried to do this by approximating the sums by definite integrals. But in the top function, only the last term matters. Likewise, in the corresponding integral, of f(x)=n^x, adjusting the top limit of the integration interval between n and n+1 results in a limit that is anywhere from zero to infinite.
@@angelmendez-rivera351 Now really, I'm sure you know what I mean and you agree with me. Lim as n->inf of n^n / [sum(n^k) from k=1 to k=n-1] is infinity, and you can use the sum the geometric series to verify that.
I tried solving this problem recently (about a month ago) and I got to around 4:24 but didn't know where to go. Dividing by n^n is very clever, I should have thought of that
I would think that it should be a limit of zero. For the top, as you add 1 to n, you increase the numerator by less than you increase the denominator. For example, the 3 values (n to the 3rd and 3 to the nth) at n=7 are 343 and 2187, but at n=8 are 512 and 6561. Thus, as n approaches infinity, the denominator grows exponentially faster than the numerator, meaning the limit of the whole is zero.
Red X Except this is simply not true. The numerator is a sum of exponential functions. This much is objectively true. It is a sum of powers, and the result post-evaluation is a function of the growth order n^n. Meanwhile, the denominator can be evaluated using Faulhaber's formula, and when you do, the highest order term is n^n as well.
Some of these commenters can't tell the difference between solving a problem and proving a theorem. They point out flaws in your logic, but not in your answer. I guess they take points off for not showing ALL your work. Too many people comment on these math videos to show off their "intellectual superiority", but end up looking like a fool.
How did you knew that dividing by n^n would do the trick? It works perfectly but damn i never would've thought that, or is just out of pure practice that you learn tricks like that? Also you should put your girlfriend playing piano on the outro again, i really loved it
6:31 why can't we conclude that the limit is inf since the numerator is approaching 1 and denominator is approaching 0 because it is a sum of infinitely small functions?
Hello Could you make a short video on how you did the limit of the denominator Cause the limit theorem (e^ab) you mentioned in the video would fail towards the end as you approach the "infinite th" terms
@10:00 The proof at that point uses (with A(i, n) = (1 - i/n)^n, for n > i >= 0) : (1) LIMIT(as n approaches infinity)[SUM(i = 0 to n -1) A(i, n))] = SUM(i = 0 to infinity)[LIMIT(as n approaches infinity) A(i, n)]. However, (1) does not hold for all possible A(i, n). For example, if A(i, n) = 1/n, for n > i >= 0, then LIMIT(as n approaches infinity)[SUM(i = 0 to n -1) A(i, n))] = LIMIT(as n approaches infinity)[1] = 1 and SUM(i = 0 to infinity)[LIMIT(as n approaches infinity) A(i, n)] = SUM(i = 0 to infinity)[0] = 0. I thank Abathur for this example (see Abathur 's comment below). Hence, it must be proven that (1) holds when A(i, n) = (1 - i/n)^n. I'm working on it. I'd appreciate seeing a proof if you have one. I'll post a proof here if I find one.
Sure, but a simple way to show the limit of 1+1/n+...+1/n^(n-1) is 1 is to use the geometric series formula (which is extremely useful and worth knowing).
The numerator could have been done in a simpler manner without the algebraic substitution: after dividing it by n^n, it renders the whole numerator a geometric series with the ratio of 1/n. And this approach would make the solution involve two geometric series (in both numerator and denominator), more elegant feeling lol
The more general formula is: x^n - y^n = (x - y)( x^(n-1) + x^(n-2)·y + x^(n-3)·y^2 + ... + x^2·y^(n-3) + x·y^(n-2) + y^(n-1) ). The terms of the second factor are all the products of powers of x and y such that the degree of the product is n-1. This way, when you multiply it by (x - y) you get a copy of the second factor with an extra x on each term and also a negative copy with an extra y on each term and now all the terms have degree n with a positive x^n, a negative y^n and all the rest cancel each other. In this case, y = 1, so it's kind of invisible what's really happening.
+delov goaz There are way too many parenthesis in my comment. That's one thing I'd like to change for the comment section. Short answer: My comment is wrong. But there is a much easier way to deal with these sorts of problems, namely the summation operator. If you collapse your sum into a summation using Σ, you can handle the sum much easier. Then, you only need to use proof by induction to proof the theorem. It even looks much cleaner. After the proof is finished, you can "decollapse" the summation. I'm not sure if that's brute-forcing in your book, at least it's easier to understand and looks better. I found a video covering the more general version of your question. Just plug in a = n and b = 1. Unfortunately, it's in german, but the mathematics is the same. The video: th-cam.com/video/vCEfp0FPIIA/w-d-xo.html
Great maths tutorial. From a watcher point of view, let me suggest that 1. you turn on some lights on the whiteboard. We are watching this on a smaller screen TV, and it's nearly impossible to read on a grey background. 2. You arrange the whole board to be in-frame. 3. Write larger. 4. Speak slower. 5. Avoid turning away from the mike and having a chat to yourself under your breath. Thanks.
I haven’t watched yet. I get (e-1)/e. I divided numerator and denominator by n^n. Numerator then goes to 1. Denominator goes to 1 + [(n-1)/n]^n + [(n-2):n]^n.... which I recognized as 1 + 1/e + 1/e^2.... which converges to 1 + 1/(e-1). So 1 /[1 + 1/(e-1)] which is (e-1)/e
When you factorised the top, why not use the formula for the sum of a geometric series? That's the expression [Sn = a(r^n-1)/r-1] you ended up with anyway.
(n-1)(n^n-1 + n^n-2 + .... 1) First the n distributes: n^n + n^n-1 + ..... n Then distribute that first negative one : -n^n-1 - n^n-2 ..... -1 n^n-1 remains, while everything else cancels nicely from the top string of numbers and bottom string.
Intuitively, *and only intuitively*, its like factoring the (n-1) from a polynomial because you know that polynomial is equal to zero when n=1. Yes, I am aware it is not a polynomial.
search up proof of sum of geometric series, then just take the n-1 to the other side. Here's a video for the proof: th-cam.com/video/5VTmlDk4ed0/w-d-xo.html
Initial idea: Use Stolz's lemma (lim of Xn/Yn where Yn is monotonic and goes to infinity = lim of (Xn+1 - Xn-1)/(Yn+1 - Yn-1)) And use the Binomial theorem to expand Xn+1 - Xn-1 and maybe some factoring trick for Yn+1 - Yn-1.... Nah, that's too complicated :-(
Hello Sir, I have some brilliant problems on geometry and algebra. should i post them in Comment and will you make videos for them according to their difficulty level?😮 Waiting.
First you should prove that the bottom sum you calculate really converge to the geometric progression, because there an infinite amount of terms. So you can’t just do the limit of the individual terms.
A counter example of the fact that you just cannot take the limit in the individual terms of an infinite sum(grows with “n”) is : Sn= 1/n+1/n+...+1/n, where it have “n” terms. It obviously converge to 1, but if you take the limit of each individual terms, it will converge to 0!
A great way to show that it really converges to the limit you show is to bound it by 2 expressions that converges to it. The firts one is using the exponential inequality e^x>= 1+x, and the other is taking the restriction of your sum to a number less than n, and that doesn’t grow with n(the number of terms). Then you can take the limit of each term and after that you take the limit of p(or to be more formal, use ε δ definition to get an inequality that will converge to the number you want)
Mateus Caracciolo That counterexample is not a valid argument. You forget that if you take the individual terms to the limit, you get 0, but you get an infinite series of 0s, which is generally not 0. So it actually very well admits the answer, it just is indeterminate.
we must do this do this carefully
Konstanty yes
Nahi
that's to separate statements.
@Konstanty Ty z Polski jesteś?
Repetition legitimazes
Repetition legitimazes
Repetition legiti-oria
There is a serious problem in the proof as others before me mentioned. With a sum whose length goes to infinity, we are not allowed to take the limit term by term. Otherwise, you can prove that 1=0 by considering
1=n*(1/n)=(1/n)+(1/n)+(1/n)+...
Taking the limit, we obtain
1=0+0+0....=0
In fact, the limit of the denominator should have been calculated using the squeeze theorem and noting that
e^(-i)>=(1-i/n)^n>=e^(-i)*(1-i^2/n)
Yinon Nahum i agree with your example. But the bottom terms aren't zeros tho.
That was just a very simple example where things go wrong. In general, the theorem says that if {a_n} approaches the limit a and {b_n} approaches the limit b, then {a_n+b_n} approaces a+b. By induction you can extend this to any *fixed* number of summands. The theorem does not allow you to do it for a *variable* number of summands
Yea, I should have mentioned more details in the video.
Yinon Nahum thanks for the feedback. I was wondering the example you gave was in the indeterminate form of inf*zero. But the one that we had in this problem wasn't in any indeterminate form.
blackpenredpen the example is well defined. Taking the limit term by term, we obtain the sum of n zeros, which is equal to n*0=0.
Regardless of the example, taking the limit term by term is allowed when one has a fixed number of summands. Doing so when the number of terms tends to infinity is not based on anything.
"We must do this do this carefully"
1:09 We must do this carefully and repeat for double-checking
: )
Lol he cut ne statted again….good video tho
This video put a smile on my face, made me laugh, and helped me fall in love with math again. Thank you, BlackPenRedPen!
Joseph Petrow yay!!!!! I am so glad to hear it?? :)
After your more recent videos, I was scared I might've outgrown this channel, but this video felt like the right difficulty. Thank you
Keenan Horrigan thanks, i will be making several videos for my calc 1 students in the summer : )
There's a slight logic error. When evaluating the bottom, you take the limit of each term as if n goes to infinity first without considering that the amount of terms is also depending on n. It's not clear that in general this kind of exchange of limit and sum works. To do this precisely, there is more care needed, though the final answer should stay the same.
Is there a counter example or sould he just precise that it is true ?
Abathur was trying to figure that out. Usually the way you do that kind of sum is by turning it into a riemann integral
Mike Montoro i just find an example with the sum from 1 to n of 1/n : always do 1 but if you goes to the limit in the term first you get 0... So your idea was really important !
Abathur Exactly
Alright so here's what I've figured out: let a_k,n=(1-k/n)^n, A_k=e^-k. We need to show the following (some sort of uniform convergence): there exists an N such that for all k and all x>0, for all n
I knew when I saw 1-1/e there would be some sort of geometric series
This is a very interesting limit problem. It’s hard to get an intuitive handle on it.
One approach:
Top is a geometric series, and is = (nⁿ⁺¹ - n)/(n-1)
Bottom is a sum of the first n, n’th powers, which can be shown to be
∑₁ⁿ jⁿ = nⁿ⁺¹/(n+1) + ½ nⁿ + O(n) O(nⁿ⁻¹) + ...
So there are n+1 terms, each of order nⁿ. This means that you get an infinite series whose terms are difficult to establish, multiplying nⁿ.
And that makes the limit look like
(n+1)(nⁿ⁺¹ - n)
-----------------
(n-1)nⁿ⁺¹[1 + (n+1)/2n + O(1)]
which in the limit, becomes 1/S, where S is that unknown series.
Another approach to modeling the denominator is to approximate it with the integral from x=½ to n+½ of xⁿ.
And that’s [(n+½)ⁿ⁺¹ - (½)ⁿ⁺¹]/(n+1) → (n+½)ⁿ⁺¹/(n+1)
So now the limit looks like
nⁿ⁺¹
----- = [n/(n+½)]ⁿ⁺¹ = 1/[1 + 1/2n]ⁿ⁺¹ → 1/e^(½) = e^(-½) ≈ 0.60653...
(n+½)ⁿ⁺¹
Of course, this is nowhere near rigorous; the interval of integration could be adjusted slightly, and the integral is after all, only an approximation.
But I think this shows that the limit is finite and finitesimal; i.e., positive and finite.
After watching: That was really cool! So my estimate was pretty close - that was unexpected.
1 - e⁻¹ = 0.63212...
Thanks, bprp!
Fred
Thank you for sharing your thoughts with me again, Mr. Fred : )
Wish you have a great summer time!
You're welcome, and thanks! Here's wishing you the same, and more!
Fred
ffggddss yay!!
You broke the question down in a way that is easy to follow and understand. Really impressive!
I'm amazed at how many times e (and pi) show up in such unexpected places! Great video!
1:10 was that on purpose?
He didn't do didn't do it.
who knows....
There was 2 clips
A glitch in the matrix
It wasn't the recordings fault, it actually happened in real life.
We must (lim [n -> ∞] (do this)^n) carefully!
I am very interested in this lesson. We must do it. Open your heart and start to learn it🙏
I have seen the comment by @Mike Montoro (and there may be others that are similar), but there is not a slight error in logic in this proof but a HUGE one.
The way that you take the bottom limit is not at all valid; you cannot take the limit term-wise when the number of terms depends on n.
As a counterexample, consider the sum 1/(n^3) + 4/(n^3) + ... + (n^2)/(n^3). By your method, if we were able to send n to infinity for each term, then each term would go to zero and the sum of all would be zero. But this is not the limit of this sum! It can be re-written as (1/n)(1/n)^2 + (1/n)(2/n)^2 +... + (1/n)(n/n)^2. From here we can see that this is a Riemann sum of a function x^2 on a partition of n intervals of equal length between 0 and 1. So sending n to infinity would send the entire sum to the integral of x^2 from 0 to 1, which is 1/3; not 0.
Daniel Juncos Your argument is valid, but the problem is that he never took the limit term wise. Both the index of summation and the terms were taken to infinity simultaneously, which is what he was supposed to do.
Daniel Juncos If he had taken the limit termwise, he would have obtained 0 as well.
Harder than I thought indeed. I thought this would be 0 as both series diverge and and imagine doing lahospital's rule over and over and the numerator should appraoch N! whereas the denominator would still have ln(n)^n(N^n) at least. I guess lahospital's rule can not work if the partial sum (the series itself)is changing. A great lesson indeed. Thank you.
I think the rule actually works correctly but the problem is taking the derivative properly is difficult. Because you're taking the derivative w.r.t. n and n is also the number of terms in the expression. On the top you can rewrite to a constant number of terms, but on the bottom it is not so straightforward.
It’s because you have assumed that it’s differentiable, when the above series isn’t, and lhopitals rule only works if you can find elementary derivatives
Cool limit. I will have to show this to my Calculus students after we learn about series.
: )
Yay!
Very logical question.takes too much time to solve before.but now got it
Great video, but I preferred it when you didn’t disclose the answer up front, but only at the end when you finished solving! Waiting for the answer is part of the fun :)
Something wrong ?
At 8.38, "the fact" : Lim(1+a/n)^bn=exp(ab) is clear when a
I have two requests for you.
1. Can you do a video about verifying whether the fourier series of a function converges uniformly?
2. Can you do something related to group theory? (e.g. finding whether given groups are isomorphic; finding composition series of a group; is a group simple? is a group solvable? What are the divisors of a given free abelian group? etc.. if these things appeal to you at all)
Not exactly a neat way, but I instead took that n^n as common on both the numerator and denominator and canceled them out.
As the last terms become 1, I checked the value of previous term which is n-1. The numerator does fine as the previous term value ends up at 1/n, which just becomes 0 when equated.
As for the denominator, the previous term (n-1) is (1-(1/n))^n, and the one before that will be (1-(2/n))^n.
As I keep checking the previous terms, the value goes off to 0 so I just leave the denominator with three previous terms and the last term (1).
Then I plug in the approximations for those values so I sort of end up with it
1/(1+0.1350+.36786+.04956)
Which happened to converge to the original answer.
Not neat one bit, but somehow converges to it.
When you take the limit in the bottom, it is NOT taking the limit of partial sums of a series - each term is changing with n, so it is not always true that you scan swap the order of limit and series sum. To do this, you need to prove (I believe) uniform convergence of the series.
Ashwin Singh The uniform convergence of the denominator is trivial though, since it is literally just Faulhaber's formula, which is well-known to have said properties. It's not like we have to prove Pythagoras' theorem everytime we use it.
@@angelmendez-rivera351 Hold up, I've been proving it for nothing this whole time???
Formally, the limit swap at 9:22 was possible due to "Tannery's theorem" - a result of the dominant convergence theorem. I guess that you dont want to talk about a little higher topics such as measure theory because most of your viewers probably aren't mathematicians, but ones that enjoy more fundamental math - and thats ok. But its not ok to make a video out of a problem you think your viewers will find interesting and provide a pretty detailed solution, even links to the theorems in the description which makes it disguised as a formal proof while you actually provides a half-wrong one. That's a misleading. You didnt have to make this a "heavy" video with measure theory - you just had to mention the theorem..
Hey, I'm not completely sure, but if Wikipedia's not lying about what Tannery's theorem is, then I assume it won't be helping here, because we can't uniformly dominate the coefficient sequences by any integrable sequence M_k.
טוב אתה
Hey I got a question. Tannery's theorem is used for interchanging an infinite limit with an infinite summation. What trips me up is the summation in this video is not quite an infinite summation but rather the top of the "sigma" is also determined by n. You see this for a split moment at 7:59. As you can see the summand depends on n but also the number of terms. Is that a problem?
Agree, simply mention what theorem might come into play suffices. What irritates me more is that some comments say this is like "well-known", or saying "the result is still correct so no problem", just wanting to show off their knowledge? There are theorems for you to mention, not like a multiplication table that everyone knows.
@@hassanakhtar7874 you can just complete the infinite sum with infinite zero terms. In the notation of the Wikipedia page, define: a_k(n) = (1-k/n)ⁿ for 0 ≤ k ≤ n-1, and a_k(n) = 0 for k ≥ n. Then for n → +∞ lim a_k(n) = e⁻ᵏ, and |a_k(n)| = a_k(n) ≤ e⁻ᵏ = M_k, which satisfies the hypothesis of the theorem, and so the trick works.
It's the best mathematics channel I've ever seen 😍 Thank you! #blackpenredpen 😊
Selma D. Thanks!!!
The limit at the bottom is first taken in each term and then at the sum limit. That requires some justification.
Ian No, he took both simultaneously. It also works because the denominator is a Riemann sum.
To complete your proof you need to show that limite of sum ( k /n ) ^n converges to sum ( exp(-k)).
First we can say that (k/n)^n = (1 - (n-k)/n) ^n . and then we can rearange the sum by doing the following change : k'=n-k
So the sum (k/n)^n = sum (1- k'/n)
Once this is done we can show that every term of the series can be majorated like that (1-k/n)^n
THE FACT is that I was rekt by this limit, lol.
Adam Kangoroo yay!!
That doremon tune 😊 my childhood 😭😭😭😭
its his childhood too probably.
If you want, an other limit n-> infinity with the expression : 1*1 + 2*2 + 3*3 + 4*4 + .... + (n-1)*(n-1) + n*n (* : power to) ; this expression at any place in the fraction (up, down)
this is just sum of squares of natural numbers
9:22 Why is it valid here to swap the limit of a sum for an infinite sum of limits?
Yeah, I don't know either. You can't do that always.
*Breathes in* I hope this is understandable in youtube writing format.
Ok this theorem is called dominant convergence theorem and is a general measure theory result (it works for series, integrals, probabilities and more).
Suppose there is a sequence a(n,m) such as a(n,m) converges to some a(m) when n alone approaches infinity. Now consider there is a sequence b(m) such as b(1)+b(2)+b(3)+... converges, but also b(m) >= |a(n,m)| for any n.
Then you can totally swap the lim and the series sign.
So you can get the full picture, a(n,m)= (1-m/n)^n
Now we get to the point: how did BPRP know he could use this theorem? Because you can absolutely use a(m) as b(m) when a(n,m) is always positive (check) and ever increasing with respect to n: (1-m/n)^n
Macieks300 When can you not do it?
@@thedarkspeedninjashadittsux
In the harmonic series, for instance.
Or in a case such as
the lim n -> ∞ [∑_{k=0}^{n} 1/n^p + 1/2^k * (∑_{i=0}^{n} 1/2^i)]
where summing the limits of each term results in two, while the limit of the whole goes to infinity for Re(p) ≤ 1, and results in ζ(p) + 2 for Re(p) > 1.
Probably because the number of terms depend on n and since n is infinite, there are an infinite number of terms so doing so should be fine.
When you divided n power n, you may put the series on denominator in integral like function A power x from zero to one.
with the top isnt it easier to factor out n^n and cancel everything on the inside of the brackets to get n^n(0+0+0+...0+1) since the 1/n^a's will approach 0 anyway, and you have 1 because n^n/n^n = 1, so the top becomes 1*n^n = n^n, and so with ur next step of dividing by n^n on the top and bottom you get the top being 1 with much less work
I am in need of a epsilon and delta proof for the last limit. You could use lemmas, proposition and anything, but I would need to see It.
oh, that Fact didn't seem very intuitive, but for the ln of small numbers (close to 1) I do recall a limit:
lim n to infinity of ln (1+a/n) = a/n, so indeed (1+a/n)^n = e^a
So the bottom sum / n^n indeed becomes the sum of all these terms that approaches the sum 1+e^-1+...
A bit more clear method of solving series would be nice, instead of using a rule like a magic wand. Ask what is needed to remove all the middle terms so the first and last term remain:
1. the series n+n^2+.. +n^n needs a factor 1-1/n or (n-1)/n to become n^n-1. bringing this factor to the other side and setting apart n^n, the sum equals n^n * n/(n-1)*(1-1/n^n), which will approach n^n * 1 in the limit.
2. the series 1+e^-1+e^-2+... needs a factor 1-1/e to become 1-e^-(n+1). ignoring this last term and bringing the factor to the other side the limit of the series becomes 1/(1-1/e).
Clever method to obtain Geometric sum series to infinity on the bottom - took me a while to figure out why you divide by n^n. :)
Can you make a video on doing sum series for sinh^n ( or any hyperbolics)? Just for the maths :)
The invisible hand There is no closed-form formula for the sum of sinh(u)^n with respect to n.
I would have never thought of that 'note'. Can someone explain his/her thought process on how to get to this note?
.what is done after 7:01 is wrong . Here is the counter example:
integral of x on [0,1] =1/n^2+2/n^2+...+n/n^2
(You can find this by using Riemann sums).
In fact al terms on the RHS of this equation foes to 0 as n goes to infinity.
Hence, we have
integral of x on[0,1] =0+0+0+...=0
Very nice!!! I like when something family happens to apear where we don´t expect!!!
Probably worth mentioning that the step at 7 minutes is dependent on n approaching infinity as the previous denominator is finite but the subsequent one(that gets summed up) is not?
What a brilliant answer!
J.J. Shank yay
If the numerator was just n^n, when you divide it by n^n, the result would still be 1. The other terms don't effect the result.
I initially tried to do this by approximating the sums by definite integrals. But in the top function, only the last term matters. Likewise, in the corresponding integral, of f(x)=n^x, adjusting the top limit of the integration interval between n and n+1 results in a limit that is anywhere from zero to infinite.
dugong369 Except that's not how that works, and it's not true that only the last term matters on the top. It's a geometric sum.
@@angelmendez-rivera351 Now really, I'm sure you know what I mean and you agree with me. Lim as n->inf of n^n / [sum(n^k) from k=1 to k=n-1] is infinity, and you can use the sum the geometric series to verify that.
I tried solving this problem recently (about a month ago) and I got to around 4:24 but didn't know where to go. Dividing by n^n is very clever, I should have thought of that
I would think that it should be a limit of zero.
For the top, as you add 1 to n, you increase the numerator by less than you increase the denominator.
For example, the 3 values (n to the 3rd and 3 to the nth) at n=7 are 343 and 2187, but at n=8 are 512 and 6561.
Thus, as n approaches infinity, the denominator grows exponentially faster than the numerator, meaning the limit of the whole is zero.
Red X Except this is simply not true. The numerator is a sum of exponential functions. This much is objectively true. It is a sum of powers, and the result post-evaluation is a function of the growth order n^n. Meanwhile, the denominator can be evaluated using Faulhaber's formula, and when you do, the highest order term is n^n as well.
Sums grow faster than their summands do.
this is one of the most beautiful series ive seen in a while , btw , why wouldn't an / bn ---> l => a1 + a2 ... + an / b1 + b2 .... bn --> l
Some of these commenters can't tell the difference between solving a problem and proving a theorem. They point out flaws in your logic, but not in your answer. I guess they take points off for not showing ALL your work. Too many people comment on these math videos to show off their "intellectual superiority", but end up looking like a fool.
You could also say that the top is an infinite GP with n terms
The top part is just the finite sum of a geo series, so you could have just used that immidiatly
Here you have taken n/n-1 tends to one but we can't take it as infinity /infinity is undefined
sandipan chakraborty Yes, you can take n/(n - 1) to have limit 1, because n/(n - 1) = 1 + 1/(n - 1), and 1/(n - 1) is 0 when n approaches infinity.
Love from India. ♥️☺️
This is really amazing!! It just poped to me now so it is never too late to watch great math😀
We must 2*("do this") carefully!
Dok Asov yup!!
How did you knew that dividing by n^n would do the trick? It works perfectly but damn i never would've thought that, or is just out of pure practice that you learn tricks like that?
Also you should put your girlfriend playing piano on the outro again, i really loved it
Because n^n is the "dominating boss" : )
And I will put the music in again.
Lmao
Alright
Because brilliant.org provides answers to the questions...
6:31 why can't we conclude that the limit is inf since the numerator is approaching 1 and denominator is approaching 0 because it is a sum of infinitely small functions?
"denominator [...] is a sum of infinitely small functions"
None of the terms in the denominator approach 0.
Hello
Could you make a short video on how you did the limit of the denominator
Cause the limit theorem (e^ab) you mentioned in the video would fail towards the end as you approach the "infinite th" terms
the last term is (1/n)^n which can be rewritten as (1 - (n-1)/n)^n
One of the very complex limit, I have seen. Thanks bprp
And finaly, what is the right demonstration of your result ? No one is written in the description.😢
@10:00 The proof at that point uses (with A(i, n) = (1 - i/n)^n, for n > i >= 0) :
(1) LIMIT(as n approaches infinity)[SUM(i = 0 to n -1) A(i, n))] =
SUM(i = 0 to infinity)[LIMIT(as n approaches infinity) A(i, n)].
However, (1) does not hold for all possible A(i, n). For example, if A(i, n) = 1/n, for n > i >= 0, then
LIMIT(as n approaches infinity)[SUM(i = 0 to n -1) A(i, n))] = LIMIT(as n approaches infinity)[1] = 1
and
SUM(i = 0 to infinity)[LIMIT(as n approaches infinity) A(i, n)] = SUM(i = 0 to infinity)[0] = 0.
I thank Abathur
for this example (see Abathur
's comment below). Hence, it must be proven that (1) holds when A(i, n) = (1 - i/n)^n. I'm working on it. I'd appreciate seeing a proof if you have one. I'll post a proof here if I find one.
If you haven't find a solution yet you can look at my comment.
For the numerator, could you not just factor out n^n to get 1+1/n+...+1/n^(n-1) which has a limit of just 1? The n^n-1 identity seems odd to memorize
Sure, but a simple way to show the limit of 1+1/n+...+1/n^(n-1) is 1 is to use the geometric series formula (which is extremely useful and worth knowing).
My brain refuses to accept any answer other than 1 because if you go one by one, top is always equal to bottom.
zlac This is not true. Just let n = 2. Then the numerator is 2 + 2^2, while the denominator 1^2 + 2^2, and those are not equal.
The numerator could have been done in a simpler manner without the algebraic substitution: after dividing it by n^n, it renders the whole numerator a geometric series with the ratio of 1/n. And this approach would make the solution involve two geometric series (in both numerator and denominator), more elegant feeling lol
Shih Yuin Chew What are you talking about? The numerator ALREADY was a geometric sum. The denominator was not.
2:03 how?
like.. i get that its true when you multiply it out, but how do you actually get that without brute-forcing a formula?
You can also think of it as a finite gemoetric series
The more general formula is: x^n - y^n = (x - y)( x^(n-1) + x^(n-2)·y + x^(n-3)·y^2 + ... + x^2·y^(n-3) + x·y^(n-2) + y^(n-1) ).
The terms of the second factor are all the products of powers of x and y such that the degree of the product is n-1.
This way, when you multiply it by (x - y) you get a copy of the second factor with an extra x on each term and also a negative copy with an extra y on each term and now all the terms have degree n with a positive x^n, a negative y^n and all the rest cancel each other.
In this case, y = 1, so it's kind of invisible what's really happening.
+delov goaz
n^n - 1 = n^n - 1^n
using a^m - b^m = (a - b) * (a + b)^(m - 1):
n^n - 1 = (n - 1) * (n + 1)^(n - 1)
binomial expansion leads to the equation:
n^n - 1 = (n - 1) * (n^(n - 1)*1^0 + n^(n - 2)*1^1 + ... + n^(1) * 1^(n - 2) + n^0 * 1^(n - 1) )
1^k = 1 for any natural number k:
n^n - 1 = (n - 1) * ( n^(n - 1) + n^(n - 2) + ... + n + 1 )
You need the binomial formulae to deduce this fact, but it is possible to prove without "brute-forcing".
Sincerely,
ZfE
thanks!
but, why are there no binomial coefficients? am i missing something?
+delov goaz
There are way too many parenthesis in my comment. That's one thing I'd like to change for the comment section.
Short answer: My comment is wrong.
But there is a much easier way to deal with these sorts of problems, namely the summation operator.
If you collapse your sum into a summation using Σ, you can handle the sum much easier. Then, you only need to use proof by induction to proof the theorem. It even looks much cleaner.
After the proof is finished, you can "decollapse" the summation. I'm not sure if that's brute-forcing in your book, at least it's easier to understand and looks better.
I found a video covering the more general version of your question. Just plug in a = n and b = 1. Unfortunately, it's in german, but the mathematics is the same.
The video: th-cam.com/video/vCEfp0FPIIA/w-d-xo.html
Great reasoning.Thanks
Great maths tutorial.
From a watcher point of view, let me suggest that
1. you turn on some lights on the whiteboard. We are watching this on a smaller screen TV, and it's nearly impossible to read on a grey background.
2. You arrange the whole board to be in-frame.
3. Write larger.
4. Speak slower.
5. Avoid turning away from the mike and having a chat to yourself under your breath.
Thanks.
In the next video could you explain about what is converge and diverge???
Doreamon music 😍😍😍 and you are so awesome 🤗
I haven’t watched yet. I get (e-1)/e.
I divided numerator and denominator by n^n. Numerator then goes to 1. Denominator goes to 1 + [(n-1)/n]^n + [(n-2):n]^n.... which I recognized as 1 + 1/e + 1/e^2.... which converges to 1 + 1/(e-1).
So 1 /[1 + 1/(e-1)] which is (e-1)/e
This video is a masterpiece
Beautiful question!
When you factorised the top, why not use the formula for the sum of a geometric series? That's the expression [Sn = a(r^n-1)/r-1] you ended up with anyway.
Kai Cooper I do that in many limits
It works perfectly
The geometric series formula is only valid for fractional values.
Shachar H The formula is always valid in the finite case, which is what Kai Cooper was referring to.
Shachar H no...it's valid for everything. Except when ratio is 1
Thank you bla
Amazing problem 🤩👍
Man, you're AMAZING!!!!!!
Okay but why does this video start with the Doraemon theme 😂😂
YAY!!!
YAY!!!!!
Yay! Severus! Yay!
Always
Very good explanation
We must do this *do this* carefully... What a good title!
Very good one
What i didnt get well is why 1^n =1 when n goes to infinity ??
It s undefined ?right? Isn't it ? 😅
No. The indeterminate form is only when the base is a function approaching 1, not straight up 1.
blackpenredpenbluepen
Naej bprpbp
YAY
So beautiful,thanks for this.
You can use the geometric series too
Max Haibara Which he did. Do people not actually pay attention to the video?
2:06 Could anyone explain to me why this is true?
(n-1)(n^n-1 + n^n-2 + .... 1)
First the n distributes:
n^n + n^n-1 + ..... n
Then distribute that first negative one :
-n^n-1 - n^n-2 ..... -1
n^n-1 remains, while everything else cancels nicely from the top string of numbers and bottom string.
Intuitively, *and only intuitively*, its like factoring the (n-1) from a polynomial because you know that polynomial is equal to zero when n=1. Yes, I am aware it is not a polynomial.
search up proof of sum of geometric series, then just take the n-1 to the other side. Here's a video for the proof:
th-cam.com/video/5VTmlDk4ed0/w-d-xo.html
Just use the geometric sum formula for r=n and it will work easily
Initial idea: Use Stolz's lemma (lim of Xn/Yn where Yn is monotonic and goes to infinity = lim of (Xn+1 - Xn-1)/(Yn+1 - Yn-1)) And use the Binomial theorem to expand Xn+1 - Xn-1 and maybe some factoring trick for Yn+1 - Yn-1....
Nah, that's too complicated :-(
Quite an interesting problem!
No one:
Him: plays doreamon tune
Bprp I have a challenge for you
Evaluate lim. n^p sin^2(n!). Whole divided by
n tending infinty. n+1. Where 0
Why did the series in the denominator become an infinite series once you divided by n^n?
Zaephou It did not become an infinite series when he divided by n. It became an infinite an series only after he took the limit n -> ♾.
Hello Sir,
I have some brilliant problems on geometry and algebra.
should i post them in Comment and will you make videos for them according to their difficulty level?😮
Waiting.
I thought you would evaluate the denominator function as a integral
limit of the sum isn't always same as sum of the limits, so it has to be proved that they are
Oh mann i miss those videos
We must do this carefully
How is the limit as x->infinity of 1^x equal 1 if 1^infinity is an indeterminate form?
Not every limit taking the form 1^infinity is 1 but this one certainly is.
1^1 = 1
1^2 = 1
1^3 = 1
...
limit of the sequence 1,1,1,... is 1.
But at the end the limit is of a sum which length aproaches infinity. Can you really take a limit term by term in this case?
Maciek300 He didn't do it term by term. He did it uniformly, with the terms approaching the limit AS WELL AS the index of summation.
Please, this kind of topics, at Brilliant, is in which session??? Calculus???
Problems of the week! : )
@@blackpenredpen thank you very much. Cheers from Brazil
First you should prove that the bottom sum you calculate really converge to the geometric progression, because there an infinite amount of terms. So you can’t just do the limit of the individual terms.
A counter example of the fact that you just cannot take the limit in the individual terms of an infinite sum(grows with “n”) is : Sn= 1/n+1/n+...+1/n, where it have “n” terms. It obviously converge to 1, but if you take the limit of each individual terms, it will converge to 0!
A great way to show that it really converges to the limit you show is to bound it by 2 expressions that converges to it. The firts one is using the exponential inequality e^x>= 1+x, and the other is taking the restriction of your sum to a number less than n, and that doesn’t grow with n(the number of terms). Then you can take the limit of each term and after that you take the limit of p(or to be more formal, use ε δ definition to get an inequality that will converge to the number you want)
Mateus Caracciolo That counterexample is not a valid argument. You forget that if you take the individual terms to the limit, you get 0, but you get an infinite series of 0s, which is generally not 0. So it actually very well admits the answer, it just is indeterminate.
You can solve numerator just like Geometric sequence
Can you please show me how to solve this problem:
Which one is bigger? 99^50+100^50 or 101^50
Ranjana Das a hint is look at 101^50 as (100+1)^50 then use the binomial theorem
0:04 is that the doraemon theme song?? BLACKPENREDPEN ANSWER ME