Every Student Should See This
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- เผยแพร่เมื่อ 16 ม.ค. 2025
- This has always been one of my absolute favorite visual proofs!
This sum 1 + 1/2 +1/3 +... is call the harmonic series. This is a great math proof showing why the harmonic series diverges! (We basically use the integral test for series)
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
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The answer is 1 in this video, not infinity or 2
1
But.. Plank length..
no
Answer is infinity
Infinity + infinity = infinity so I guess the answer is at least 2
nah it's like at least 7
It's at least TREE(G(64))
@@lyrimetacurl0 ^Rayos numbers
@@romi5073 eulers number
I feel like it's more than 1 but never 2
Year 2, month 7, day 23: I'm starting to think this may be a loop
You can't be more wrong my friend
You see the red bar underneath
@@maxz69 bruh it was a joke
@@that1dudem I know but it's not as if the bar isnt visible. It isn't a joke if it's obvious that it isn't a joke
@@maxz69 🤓
In math, we have pretty much committed to memory: harmonic series always diverges
I like how things like 1+2+4+8+... "make sense" when looking at the analytic continuation of the Riemann Zeta function, but even then the harmonic series still diverges. It really can't catch a break
If p
CosineX yeah, but why? (the videos shows the reason about this divergence)
@@CasanovaLucas By the integral test. For a sum 1/n^p with p=1 in this case (although it applies to any value of p1, the antiderivative still has some p in the denominator so the series converges.
@@matei5929 that was a rethoric question my friend hahahaha
All infinities are great. But some infinities are greater than others.
- George Orwell
George Cantor
Not all infinites are created equal
@@sreekanthhere Oh wow- actually, I was just making that stuff up after George Orwell's quote in Animal Farm "All animals are equal, but some animals are more equal than others." I didn't realize until now that George Cantor had said the same thing 150 years ago. Haha- thanks for letting me know.
One infinite good, two infinite better
@@sreekanthhere “nuh-uh” - L.E.J Brouwer
More elegant proof:
1+
1/2+
1/3+1/4+ (sum of those 2 is larger than 1/2)
1/5+1/6+1/7+1/8 +(sum of those 4 is larger than 1/2)
...
You can keep on adding further "chains" (length 8, 16, 32 and so on), each with value of at least 1/2 (pretty obvious). This gives you an infinite number of 1/2s. This is infinity.
This is better than the clip IMHO. Nice one.
Yeah this made more sense than the vid
Thats not a proof, thats guessing .
@@theforbiddenfruit2300the video needs atleast some calculus knowledge to make sence
@@EssedikAbdelRahimStudentBachou no, it's called Cauchy condensation criterion
Ah yes, the endless truth of calculus: infinitely many things that are infinitely small giving you an exact solution to a problem…
Ahhh it's relieving to know that infinity-infinity can equal anywhere between negative infinity and positive infinity. Wait no. That stresses me out. Why calculus! WHY!!!
The video literally shows you that no, it doesn't always give you an exact solution. No, infinity isn't an "exact solution".
@@zlosliwa_menda Congratulations!!! _Matthew Klepadlo_ has introduced you to *Sarcasm* yayyy! 🥳👏
@@e.Kab. Why would he be sarcastic about this?
It gives you a very close approximation.
+ You just can't add infinite numbers
- Hold my integral
Series too
How many times are you gonna ask me bro😭😭i wanna go home
fr bro is playing with us😭
Infinite amount of times obviously
Bro stuck in loop 😢
😅
😂😂😂
Yes, and if you rotate that graph around the x-axis the volume will be π.
π ∫∞₁(1 / x²)dx = π(−(1/∞)+ 1⁄1) = π(0 + 1) = π
math is all patterns that make no sense but it’s cool
@Luke-lj1hb accurate
Yeah you’re not just rotating it, squaring the denominator makes the series converge faster which leads to its area being finite
This phenomenon is known as the Gabriels Horn, whilst having a finite volume, its surface area is infinite
For the next person to say the answer is ~2, that only applies to a sequence where the bottom number doubles every time. In this sequence it goes over two right after 1/3.
@seanclemson9554 supertasks are a philosophical thing, not really mathematical. I guess you could call it mathematical but usually supertasks are contradictory or paradoxical like the grim reaper supertask.
Honestly didn't believe you but yeah, it's true.
1+0.5=1.5
1.5+0.25=1.75
1.75+0.33=2.08
Does make me wonder though, when does it pass to 3? How much longer does it take to pass to the next whole number each time?
@@pineappleudh6561
As 1/3 > 1/4 > ... the infinite sum can be replaced by the smaller sum:
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + ...
= 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + ...
= 1 + 1/2 + 1/2 + 1/2 + ....
Every extra 2^k terms adding an extra 1/2 for k=0, 1, ...
So after 1 term the sum is 1
K=0, so add 1/2
So after 1 + 2^0 = 2 terms the sum is 1.5
After another k=1 terms;
ie after 1+ 2^0 + 2^1 = 4 terms the sum is at least 2
After another k=2 terms,
ie after 4 + 2^2 = 8 terms the sum is at least 2.5
And after another k=3 terms
ie after 8+2^3 = 16 terms the sum is at least 3
And so on. Note that the sum being larger will reach the next integer before this smaller sum, so it gives an upper bound on the number of terms required.
@@pineappleudh6561 it passes 3 at x=11. let I(x) be a number that follows sum I(x) times n=1 (1/n)>n, then I(x+1)=D(x) that follows sum D(x)-I(x) n=I(x) (1/n)>n+1, it's not that easy to find D(x)
@@pineappleudh6561 one thing, that almost all I(x) is p_(p_n), and I(x) that doesn't follow it it I(1),I(2),I(6), and all three I(x) i found doesn't following that rule, is that, and all three x is factorials. 1!,2!,3!, and numbers under factorials are 1,2,3.
“put the excess area in a cap and call it Mascheroni”
You're yanking my doodle!
Can you prove is it rational or irrational
I have a truly marvelous demonstration that it is irrational that this TH-cam comment is too narrow to contain.
@@gytoser801 You really think that can be answered right here right now huh
*”Euler-Mascheroni”, but then again, if we apply this all over mathematics, every constant and theorem would be “Euler-Someone Else”
No comment about the perfect loop?
whoa
i was looking for one
Yeaa
I was about to watch it for the 2nd time lol
It's been long enough for people not to care about this anymore
@@joetrident its the only one which caught me off in a while now, it deserves my respect at least
“Just under two” me every time someone asks if I know my limit.
Then, the limit is 2, not "just under two".
What limit did you have in mind when making this comment?
It's called harmonic series and with n approaching infinity the sum will also be infinite. You can find a nice proof of it on wikipedia
That one math teacher who doesn't leave you until you get the answer:
The result is so harmonic
Nice one 👍
crazy how the format of this video made the loop so seamless!!
He is idiot
If you change the series by rounding each fraction down to the nearest 1/x fraction, where x is a power of 2, you'll see that this series is the same as repeatedly adding 1/2 (per the grouping in brackets I've made below). Since this clearly smaller series, can be shown to be equivalent to repeatedly adding 1/2, and thus equal to infinity, the original larger series must be infinity as well.
1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 +1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + ...
That's the classical argument due to Nicolas Oresme - clearly the more elementary proof than using an improper integral.
I might be confused, but overall you cannot rearange terms in non-absolutely convergent series, the result will change.
@@babyboy5553 I rearranged nothing. I simply rounded every term down to the nearest power of two reciprocal, meaning my series is smaller. The brackets are solely to show that those powers of two can be grouped up and thought of as repeatedly adding 1/2, to make it easier to see that my smaller series is equal to infinity. They are not required for that statement about my series to be true, nor are they meant to be a mathematical part of my smaller series. The grouping is explanatory only.
@@babyboy55531) No rearrangement here, just an application of the _direct comparison test:_ the series written down by @TurkishKS is divergent and its terms are always less or equal than the corresponding term in the harmonic series, which hence must divergent all the more.
2) Even a rearrangement would have done no harm: series with nonnegative terms can be rearranged at will without affecting their convergence behaviour, because they converge if and only if they converge absolutely. (That theorem of Riemann you are thinking of deals with convergent, but not absolutely convergent series; those must have infinitely many positive and negative terms.)
integral of 1/x = ln|X|=> which means that as x tends to infinity, the limit will also be infinity
Great representation. For anyone curious, this is know amongst mathematicians as the harmonic series.
WTH! That loop is flawless!
Mom the video isn’t ending
Thank you so much! Bri and keep making videos like this. Almost nobody makes concepts such easier.......... 👍❤
you are gay?
That was a butter smooth loop
also u can do without integrals
1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8...>
v v v v v v v v
1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16...=
1/2+1/2+1/2+1/2+1/2+1/2+1/2+1/2+...=inf/2=inf
This is how the integral test should be taught in school. This is so intuitive!
It is taught like this tho
@@-_-_-_-_ it i taught, but some dont listen. Or some cases the trachers just explain ot like shit
Most likely the former
@@-_-_-_-_ Maybe your teacher did it that way, but you can't assume everyone else had a teacher like that
@@gabedarrett1301 that is calculus, it was taught in last year of high school to me, its litterally how it is defined.
@@ha-kx9we Well good for you, but again, you cannot assume my professor taught it intuitively like this
No calculus needed. Group up the terms like this: the first term, then the second, then the next 2, then the next 4, then the next 8. The first is greater than 1/2. The second is greater than 1/2. The next 2 are each greater than 1/4, so their sum is greater than 1/2. The next 4 are each greater than 1/8, so their sum is greater than 1/2. The next 8 are each greater than 1/16, so their sum is greater than 1/2, and so on... You get the sum of infinitely many times 1/2 which is infinity
Yet 1-1/2+1/3-1/4+…=ln(2). Fascinating
Very good video I can see you put the work in!
Its the series/integral equivalence theorem.
See: Integral test for convergence on Wikipedia.
you are gay?
me who thought the answer was 2
That's 1/1 + 1/2 + 1/4 + 1/8 ....
X below is divided by 2
@@fos1451 sweet. Thanks 👍🏽
@@fos1451 does it not work with 1/1+1/2+1/3?
i kept going further and further on the calc and it did seem to just get closer to 2
@@georgiaoni6101
The answer would be bigger than 2, are you sure you put it correctly? It shouldn't be "closer" to 2, it should be bigger than 2
I just tried it and you already got the 2.08 on 1/4 and 3.01 on 1/11
@@fos1451 ah no i did it wrong i was doing 1/2 1/4 1/8 oops
them: “so NOW what if i ask you this question”
me: i still don’t get it
edit: guys ur smart but i barely passed seventh grade math so the chances of me ever understanding this are extremely low 😂
The point is that any series of numbers repeated infinite times, if not inversely exponential and not equal to zero, no matter how small it seems to be, is equal to infinity
@@Matt-wh6wj i can't get it
for me that looks like a 1,5 if u use rectangle
@@Matt-wh6wj have you heard of p-series lol
@@Matt-wh6wj bro if you have something like an infinite gp with r
@@hellfireofdooom8876 the p series is a divergent series
the series diverges via the integral tests
as the pseries rule states, if n > 1, the series converges and if its 1 or less then it diverges
1/n limit approaches zero but proven by the integral test it diverges
Its also known as the harmonic series. It does goes to infinity but hella slow.
I like how the question is asked infinitely
This is amazing.
do the wallis product sometime around :)
My nigga💀
@@jordan7828 Hello Chris.
Yes, but if you add 1/n^2 it's actually finite
I hate the harmonic series so much because I know it diverges but it grows SOOOOOOOOOOOOOOO SLOWLY
YOU JUST EXPLAINED THIS BETTER THAN ANYTHING I'VE EVER FOUND. OH MY GOD THANK YOU SOOOOO MUUCHHHH 🤩🤩🤩🤩
If U subtract we get Euler Mascheroni Constant 😊
when I first heard about this they proved it by using a series that goes 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 ... and if you compare it to the original series
1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 ...
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 ...
you can see the original series is clearly larger
and you can also see the smaller series is basically just repeatedly adding 1/2 and if you add the same number over and over again it will eventually escape to infinity
and since the original series is larger it will also escape to infinity
Honestly, that method is so much more straightforward than "develop enough calculus to know that the integral of 1/x grows without bound, and then compare areas"
Yes, that was by Nicolas ohrem, French mathematician.. (idk if I got the spelling right)
I was surprised that 1/2^n sum is the same answer as 1/triangle numbers sum, even though the triangle numbers one is not exponential 🤔
@@lyrimetacurl0 hmm true, that is quite interesting... yes the triangle number problem can be done by partial fraction decomposition
@@diribigal This method is cool but not generalisable. The calc method is not as cool but very easy and very generalisable.
His loop is perfect 🫡💪
Yep, it diverges. Thank Professor jerrison for teaching me that in MIT 18.01.
From coder's perspective: for each 10^n the sum will be by around 2,3 higher
10: 2,93
100: 5,19 | ~+2,3
1k: 7,49 | ~+2,3
10k: 9,79
100k: 12,09
1M: 14,39
Each of them is ~ +2,3 more than last 10th's power sequence.
Script:
var number=0;
for(var i=0;i
Thanks for the code dude can u also do python
@@KrishnaPahuja8 \t means Tab
number=0
i=0
while(i
Another good explanation for this problem is this:
Take this equation, for example
X = 1/9 + 1/10 + 1/11 ..... 1/16
All of these are greater than or equal to 1/16, so let's just say x = 1/2 for demonstration purposes
Repeat this with 1/17...1/32, 1/33...1/64, and so on to get infinity because ½ × infinity is still infinity
Sorry if that didn't make sense
S = 1 + 1/2 + 1/3 + ....
√S² = √(1 + 1/2 + 1/3 + ...)²
S = √(1 + 1/2 + 1/3 + ...)²
S = √(1 + 1 + 1/6 + ...)
S = √2 + 1/2
S = √4/2
S = √4/√2
S = √2
S = 1.414213562373095...
Now this is - 1⁄12
I thought it was -12
That has been proved wrong, it was never true, it doesn't even make sense
Clickbait and that was 1/n^2
@@rawenmakboul7679 Where?
@@ravenking2458 here is the link
You start with 1,after 3 sums you got a 2.08. So after many many sums you keep getting bigger numbers until you reach infinity.
now subtract the area from the rectangles and you get 0.577. Math is awesome
This is what my calculus teacher told us:
1/1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8... = 1 + 1/2 + 1/2 + 1/2 ...
If we reduce everything so that the denominator is a power of two (this is less than this series), it will continuously add 1/2, and because 1/2 * infinity = infinity. Because the power of two series is divergent, the harmonic series must be too.
This is the solution that people who haven't finished high school still have the tools to understand.
@@MasterHigure or because it’s an other way to see the problem.
Was going to comment this. Much more simpler than involving calculus
@@MasterHigure lol are you high on yourself ? :-) ridiculing his method saying it's for high school kids
@@AalapShah12297 Yeah, that is what I was thinking. And I myself am actually a Cal B student right now and we learned improper integrals and how the harmonic series is divergent within a month of each other. So the only people who would know that the integration of 1/x from 1 to infinity is divergent and not know about the harmonic series being divergent would be a very small group of people.
My favorite explanation for why the harmonic series diverges is this: we can collect every 2^n terms and observe that their sum is at least 1/2. For example:
1 > 1/2
1/2 >= 1/2
1/3 + 1/4 > 1/4 + 1/4 = 1/2
1/5 + 1/6 + 1/7 + 1/8 > 4 * 1/8 = 1/2
So the sum of the whole series is at least 1/2 + 1/2 + 1/2 + …, which clearly diverges.
I end up confused, just like in class
Sum(1/n) is a Dirichlet series with exponent=1, therefore is divergent which means its sum its +inf
the loop was so clean he be asking the same question infinite times.. 💀
The reasoning seems obvious when it's told you well like this, but I struggled a lot with this one haha.
QUESTION: why was the variable t introduced in the final expression? Shouldn't this just be x, with x tending to infinity?
Since the function 1/x already uses x, t was used instead to occupy the limit
An extract taken from the introduction of one of Euler's most celebrated papers, "De summis serierum reciprocarum" [On the sums of series of reciprocals]: I have recently found, quite unexpectedly, an elegant expression for the entire sum of this series 1 + 1/4 + 1/9 + 1/16 + etc., which depends on the quadrature of the circle, so that if the true sum of this series is obtained, from it at once the quadrature of the circle follows. Namely, I have found that the sum of this series is a sixth part of the square of the perimeter of the circle whose diameter is 1; or by putting the sum of this series equal to s, it has the ratio sqrt(6) multiplied by s to 1 of the perimeter to the diameter. I will soon show that the sum of this series to be approximately 1.644934066842264364; and from multiplying this number by six, and then taking the square root, the number 3.141592653589793238 is indeed produced, which expresses the perimeter of a circle whose diameter is 1. Following again the same steps by which I had arrived at this sum, I have discovered that the sum of the series 1 + 1/16 + 1/81 + 1/256 + 1/625 + etc. also depends on the quadrature of the circle. Namely, the sum of this multiplied by 90 gives the biquadrate (fourth power) of the circumference of the perimeter of a circle whose diameter is 1. And by similar reasoning I have likewise been able to determine the sums of the subsequent series in which the exponents are even numbers.
Im in grade 10, and i can barley do math 7 grades below my level, but this was cool.
From which country you are ?
Not Indian Ig
can also be expressed as integral 1+x+x^2+...+x^n from 0 to 1 and n tending to infinity. solve the limit and we get infinity :)
You can prove it’s infinity due to the p series tests for sums, here it is 1/n^p where p is 1. In order for the sum to converge (not be an infinity) p must be greater than 1. Here it will diverge to infinity, which has already been proven to great lengths about this one (the harmonic series) but it’s still fun to prove yourself
Another way that is coming to my mind: if we write it as
∫(1 + x + x^2 + x^3...)dx from x=0 to x=1 , we get the same series, ie 1+1/2+1/3...
solving, =∫ (1)/(1-x) from x=0 to x=1 (using infinite GP formula, and here common difference
This was literally like “do the integral test lmao”
People: Perfect loop doesn't exist
This short video:
There are way easier proofs, for example you can easily figure out that you can divide it in an infinite amount of areas with a size bigger than one half
It’s interesting because you’d think when you take the limit you get 1/infinity (which is 0) but the infinite series of 1/x is divergent; it never converges to a finite value. It’s also known as a harmonic series.
Conversely though the alternating harmonic series is convergent; it converges to a finite value.
It clearly sums to -12.
No dude, it's 1+2+3... = -1/12(Atleast that's what Ramanujan stated)
@@ravenking2458 ramanujan stated that for 1 + 2 + 3 ……..
this is 1/1 +1/2 +1/3 ………..
@@AdityaGupta-qk5ee yep, sorry i forgot to mention that. I meant that he probably did reciprocal of the whole series of 1+1/2+1/3....
The reciprocal of a series isn't the series formed from the reciprocals of the terms.
1+2+3+4...=-1/12 (if you want to give it any value)
but
1+1/2+1/3+1/4...=infinity (if you want to give it any value)
@@RaRa-eu9mw Dude i said the same thing, he probably did the whole reciprocal of the series which was obviously wrong
The loop is infinite ♾
you are gay?
Help stepmathbrother, I'm stuck in a loop
now that integral subtract from the series and you get γ (the euler mascadoni constant) =~0.577
There is proof that doesn't require integrals, which makes it more elegant IMO.
You can lower the sum by replacing some of the terms with smaller terms to write it as:
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + ...
which doesn't seem so clever at first because I just substituted each denominator with the next power of 2.
However, when you go to calculate this new sum, which is smaller than the original sum, you get:
1 + 1/2 + 2/4 + 4/8 + ... = 1 + 1/2 + 1/2 + 1/2 + ...
and it much more obviously diverges, and means the original sum diverges as well.
As a bonus, you can see the divergence of the new sum is logarithmic. If you think about it, each time we need twice as many terms to increase the total sum by 1/2, which is a logarithmic behavior.
I had a teacher who explained this to me something like: ”1 > 1/2, and 1/2 is >= than 1/2 (I mean it’s equal to but that’s not the point) and then (1/3 + 1/4) > 1/2 and after that (1/5 + 1/6 + 1/7 + 1/8) > 1/2 and you can go on and on like this and add more and more numbers that are greater than one half.” Then I don’t remember the formula but he wrote it with summary notation that you get ”n halves” I think and so the series is indeed infinite if you keep adding numbers that sum to one half (or greater).
Well in set of real numbers you can always find one which is smaller then the previous, so nothing magical here.
That's how I learned. I guess it's the best explanation for showing the sum is infinite 'before learning the calculus'.
1/1 + 1/2 + 1/3 •••• > A: 1/1 + 1/2 + 1/4 +1/4 + 1/8+ •••
A= 1/1 + {1/2+ 1/4*2} + {1/8*4+1/16*8} + •••••
A= 1 + 1 + 1 + 1 + •••••
=infinity
-From Korean student
Omg ty
Btw I don't understand the relation between area under the curve and rectangle with the function. Is it valid for every sum?
how is the area of the rectangles represent the sum?
is it because its the same as sum of 1/x ×1 from x to inf?
The first rectangle is defined with an area of 1, the second is half its height so 1/2, the third a third of the first rectangle's height so 1/3 etc
Yes
Look at it this way: the area of each rectangle is simply the rectangle height (1/n) times the base. Because we're using indices going 1,2,3,4 etc we're actually looking at a base of width 1. Adding up all those fraction is this equivalent to adding up those areas.
With the help of heron formula calcius pytogoras theroem and newton law of intertia combined with integration we can se it more than 1
Also, if you round down 1/3 to 1/4, as well as rounding down 1/5, 1/6, and 1/7 to 1/8, and so on, then if you add those up it's 1 + 1/2 + 1/2 + 1/2... which is infinite
There is an explanation without using calculus (Cauchy's integral convergence test for sum of series):
Let's introduce new series with elements each not greater than the respective initial one, precisely:
{1; 1/2; 1/4; 1/4; 1/8; 1/8; 1/8; 1;8 ...}
Now let's look at the sum of this series:
1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+... = 1 + 0.5 + 0.5 + 0.5 ...
As we see we are constantly adding 0.5 to 1, as the sequence is infinite, the process lasts infinitely long. So the sum goes to infinity and beyond, hence this series diverges. As the initial series has elements each not less than this one, then it diverges too, so it's sum goes up to infinity too.
Thats the proof we used in our real analysis class
There is a much easier way to see why it is infinity. We can substitute each term of the sum with 1 over the next power of 2. For example, 1/3 we change it by 1/(2^2)=1/4. Therefore, we go from 1+1/2+1/3+1/4+1/5... to 1+1/2+1/4+1/4+1/8+... affirming that the original series is greater than the second one, as 1/3>1/4, 1/5>1/8... and so on. From the second series we can group terms this way: 1+1/2+2*1/4+4*1/8+8*1/16+... = 1+1/2+1/2+1/2+1/2+... = 1+2*1/2+2*1/2+...=1+1+1+... which means that, given the series is infinite, we are adding up infinite ones. Therefore, as our original series is greater than the second one, and the second one is infinite, the original is also infinite.
To be very precise, it is totally wrong comparison. Actually area under the curves give us the total sum area of continuous function that is 1/x here but we don't have x= 1/3.3 or x=1/4.2 etc. in the domain here in the question. Because our domain is restricted at integers in the denominator. So its not continuous.
You totally misunderstood the explanation.
1/1 + 1/2 +1/3 .... Is just area of rectangles of breadth 1 and length 1,1/2,1/3,1/4......
We are comparing the sum of areas of discrete rectangles and the integral of the continuous function 1/x.
Our goal is to proof that the sum of 1/n (harmonic sum) is divergent.
The thing is the harmonic sum is totally the sul of the areas of the rectangles.
How to construct these rectangles :
The begin of a rectangle is an integer and the end is next one. So the width is one.
We made them so as to the height of them is f(n) = 1/n
Then, the area of one rectangle is width * height = 1 * 1/n = 1/n
Then the sum is the harmonic sum
which is by definotion the sum of 1/n
The second area, the blue one , is the integral of 1/x
which is ln(x) , given by the fundamental theorem of calculus. This function diverges to infinoty so the blue area diverges to infinity.
The area of the rectangles is GREATER than the area under the curve. so we conclude that the area of the rectangles is also infinity.
*To conclude, we use a continous function to calculate a discrete sum because we know how to deal with functions,integrals...*
Yes infinity. If this taken upto n terms, the sum will be approximately log(n)
I just knew it as 1/3 -> 1/4 1/5 ->1/8 etc so you get 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8…. The quarters become a half the eighths become a half as well so you get 1+ 1/2+1/2+1/2…. This is smaller than that so you know it’s infinite
Your so smart
This can be understood from the Machlaurian series expansion of Log(1-x) .. putting x = 1 which gives the answer infinity.
Ahh yes, the harmonic series :)
Its infinite , that's like , that elephants father can fit in the shade, and that, elephant is big how big it's the biggest , but then what about his father he is bigger than that biggest elephant , so do we have have enough shade space to fit the elephant 🙂😆
you could prove ln(x)-h(x)=v as x approaces infinity which makes it trivial
What's funny though, is that if you revolve that area around the x axis and take the volume, you get a finite area (that is a multiple of pi if i remember correctly). Just another reason infinity isn't a number
You don't need calculus.
1/3+1/4 > 1/4 + 1/4 = 1/2
1/5+1/6+1/7+1/8 > 1/8+1/8+1/8+1/8 = 1/2
and so on
The sum > 1+1/2+1/2+1/2...
= Infinity
The answer to “1/2+1/3+1/4….” Is 1, but what about “1/1+1/2+1/3….”? That’s 2 (Zeno’s Dichotomy Paradox)
Oh really? I thought 1/2+1/3+1/4= 6/12+4/12+3/12= 13/12 which a guess is bigger than 1
Wut the hell do you mean
@@itismethatguy since we’re working with infinity, smaller and smaller numbers get closer to one
@@YeahEsCereal thats just wrong lol
1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... is equal to infinity not 1. In the denominator, we have all the natural numbers, and this serie is called "Harmonic serie". This one diverges as the video proved.
The serie which is actually equal to 1 is the geometric serie
1 + 1/2 + 1/4 + 1/8 + 1/16 ... = 1
Here we only have the powers of 2 in the denominators
This serie converges as |1/2| < 1
The terms for the two series are smaller and smaller (converge to 0) but the terms for the geometric serie converge faster than the harmonic one.
Here’s best proof for this which doesn’t involve calculus. Begin with the sum:
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + …
We can compare it to a different sum:
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + …
>
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + ….
Because of how each term increases to the next, we know the second infinite sum is less than the first infinite sum if both have integer solutions. This means that if we can prove that the second sum evaluates to infinity, then we also prove that the first sum also evaluates to infinity.
We can simplify the second sum as follows:
Sum = 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + ….
Sum = 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ….
Sum = 1 + 1/2 + 1/2 + 1/2 + …
Thus, the second sum is an infinite sum of 1/2. which is infinity. Because the second sum is equal to infinity, the first sum 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + … = infinity.
Calculus is way faster
Using Calculas.
Using expansion of ln(1-x) = -(1/1 + x^2/2 + x^3/3 + x^4/4 ....... )
Putting x=1
Ln(0) = -(1/1 + 1/2 + 1/3 +..........)
Therefore,
-Ln(0) = 1/1 + 1/2 + 1/3 +.....
Inf = 1/1 + 1/2 + 1/3 +.....
You need to separately show ln(0) is this sum. We only know the radius of convergence of the taylor series of ln(1-x) is 1; we don't know about the end points without checking them explicitly.
need more pages like yours on social medias
I think the proof by grouping values until they are larger than 1/2 is easier to understand.
This is stupid. The curve is one representation of the function, and the rectangles are another. To suggest that they can be superimposed on one another and then calculate what is no more than a representative visual difference is stupid. Congratulations on fooling people
Thanks. I was worried, that nobody mentioned this.
Technically the difference is the rectangles show the area of a sequence were they keep adding fractions with an integer value, where as the curve is an addition where denominators differ by an infinitely small number each, time so there is technically a difference.
They are not so different representations.
The rectangles are constructed so that the height of the rectangles is the same as the value of f(x) where x is an integer.
The area of one rectangle is the product of its width with its height.
Its width is n+1 - n = 1 (The beginning of the rectangle is on an integer and its end is on the next one)
Its height, 1/n
So the area is 1/n
The harmonic sum is the sum of 1/n that is to say that the harmonic sum is of the areas of all rectangles.
And the blue area is the integral of the function 1/x.
This is how we do an Integration test for convergence.
en.m.wikipedia.org/wiki/Integral_test_for_convergence
Nothing so surprising... .
1+0.5+0.3(point three repeating)+0.25>2,so at least 2
By the same logic we can say that adding the first e^n terms of the series (we take the smallest integer greater than e^n ) and it will always be greater than n by the same logic of area...
Thus the infinite sum is larger than any finite number and thus ♾️♾️
The sequence (s_n):s_n=1+...+1/n is not a Cuachy sequence since s_(2 n) - s_n>=1/2. Hence, (s_n) doesn't converge. Since is increasing, s_n must go to infinity.
Sum of an infinite harmonic series
Writing "infinite" is redundant, because the harmonic series is an infinite series.
Harmonic series
My brain after watching: WHAT THE F*CK *EXPLODES*
This definitely is one of the answers of all time
The best way to understand that this is a divergent sum is to compare 1/2 to 1/3 +1/4 since 1/4 + 1/4 = 1/2... and you can keep going on doing this for every group of 1/2^n sub sums and you will always get + 1/2... even if you may need to sum many and many elements.
E.g.: 1/5 + 1/6+ 1/7+ 1/8 > 1/8 * 4 which is equal to 1/2... and sum of 1/9 up to 1/16 again is bigger than 8×1/16 that is the sameof 1/2... so no matter what you can keep adding up terms up to infinite
So no computers and calculus needed... just good logic