A nice approach to the alternating harmonic series
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- เผยแพร่เมื่อ 13 ก.ย. 2022
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Very nice. I was very worried for a moment that we were reordering the summands of a not-absolutely converging series, but of course we just did that to calculated the limit of the partial sums, not actually change the partial sums.
His initial remark about the convergence of the series is essential. Because it is convergent, we know that the limit of the series equals the limit of the even partial sums. Otherwise we would not have calculated the limit of the series, but of a related series. Consider
(1-1)+(1-1)+(1-1)+….=0. Without the parenthesis, the series diverges of course.
@@yunoewig3095 I am referencing the fact that reorderings of the alternating sum can converge to a different number (or even diverge).
@@zairaner1489 If all cycles in a permutation are finite, then the sum will be invariant under that permutation.
@@yunoewig3095 or equivalently, if the permutation only permutes finitely many terms, which makes sense since the tail is left undisturbed.
@@justanotherman1114 In fact, what he is doing here is not even a permutation, he is straight up splitting terms, so it's a more complex operation than that.
After watching your videos for about a year I'm impressed with the quality and breadth of the material you present.
This was very interesting!! Thanks Prof Penn
Thank you, Michael! I hadn't predicted the way of solving up to the end of the calculation.
Sweet video, very neat. Thanks professor!
Got my sweatshirt yesterday too, its awesome. Thank you!
Another cute solution picks up where you split S_2N into a difference of harmonic series. Let H_N = 1/1 + 1/2 + ... + 1/N be the harmonic series. Then you had that S_2N = H_2N - H_N. If you know the fact that lim N->inf H_N - ln(N) = C (a constant), then S_2N = H_2N - H_N = ( H_2N - ln(2N) ) - ( H_N - ln(N) - ln(2) ). Taking limits as N -> inf we get that lim S_2N = C - C + ln(2) = ln(2) :)
I would imagine this would be the way they teach it in undergraduate courses...
This closed form of the difference of partial sums H_{bN} - H_N (b \in positive integers) was also explored in a collaboration integral of floor functions to Riemann sum video with @blackpenredpen
Love your explanations Professor!^.^
Thank you, professor!
That was great!
Great video!
Wonderful!
رائع كالعادة.شكرا لك أستاذ
Very nice!
great job mst michael
9:17
Ok, great!
Maybe we can do similarly with the alternating harmonic series of the odd numbers, and with the function arctan.
Very Nice
I'd like to see a double (or triple or higher summation) of (-1)^(m+n)/sqrt(n^2 + m^2).
This is related to the Madeulng constant. I don't think there is a closed form for it, or none that I've seen.
Nice！
You can also use ln(1+x) = sum (-1)^{n+1}*x^n/n, but that would be overkill
Or, as blackpenredpen would say, "use the best friend" to get the sum (-1)^n * x^(n+1)/n+1) (via integrating the geometric series)
That's probably easier.
I don't see how you can use this without knowing a priori that it converges to ln(2) at x=1, which is exactly what we are trying to show to begin with.
Even if we take as a given a strong result like ln is holomorphic on appropriate domains, we still have the problem that ln(z) has a pole at z=0. Therefore the radius of convergence of the Taylor series centred at 1 (which is the formula you've used) is exactly 1. In general, Taylor series may or may not converge to the true value of the function, or at all, at the radius of convergence. So with this alone, we don't get any information on the convergence at x=1.
For this to work, you would have to additionally know that the series does in fact converge to ln(2) at x=1. But as I say, this is what we are trying to prove to begin with, so this isn't a fact we should be able to use.
This is not overkill, it's the standard way
@@stanleydodds9 I don't get what you mean, if we have enough knowledge to derive taylor's theorem (which idts requires knowing this before using?) then what's the problem?
GREAT
I'm lost ! I have to read again about Rieman' series.
This series is a bit trivial if you can recognize the power series representation of ln(1+x), good proof though.
That's what I love Math
🐐
What about Riemann’s rearrangement theorem? Don’t you have to also specify an order of the terms?
You did pick an order by the choice of partial sums, just did not emphasize that? Very nice development though.
@@glennjohnson4919 Yeah, what he's reordering is a finite sum, which doesn't change anything, except that the resulting expression doesn't have the form of a limit of partial sums (so it can't be turned into a reordered infinite sum).
This was pretty cool! Could you make a video about the sum of a harmonic progression to n? Like an entire analysis kind of thing of 1/a, 1/(a+d)...? I was confused about this topic and don't seem to understand any sources or find reliable ones.
That second one needs to be 1/(a + d).
@@forcelifeforce Sorry that's exactly what I meant, was in a bit of a hurry lol
Maravilha
Thankyou
I have a degree in math but cant find anything to do with it
Your videos help me keep my edge sharp
great...w Torricelli
By "the terms go to zero" you mean that they decrease in absolute value, not just that they converge to zero in case anyone else was also confused about that.
I think there is no difference between converging to zero and decreasing in absolute value
@@Senteggo It's been a while, but I think I meant that the absolute values form a decreasing sequence, while a convergent sequence can sometimes take steps away from 0.
@@iabervon can you give an example?
@@Senteggo 1/(2^(n+(-1)^n)), which goes 1, 1/8, 1/4, 1/32, 1/16, 1/128, 1/64, ... Half of the terms are larger than the previous term, but the series converges to 0.
@@iabervon wolfram alpha says it diverges
H_2n - H_n has limit log(2n)+gamma - log(n) -gamma =log(2n)-log(n)=log(2). Can stop as soon as you get there
The answer ~.69 which is nice
cute!
I get nervous when such tactics are used. It seems-at first glance at least-that an advanced technique-calculus-is being used to build one of its own foundations-series sum.
It's like supporting the third floor of your building by using cables hanging from the sixth floor.
It's like using the distance formula to prove the Pythagoras Theorem. Or the Maclaurin Series for finding the power series of _e._ Or algebra to prove that 1+1=2. All of which-by the way-my students are guilty of doing.
Now you can do all that in the Math Olympiad and similar contests, where you're allowed to use any "well-known" result to solve the problem at hand. But in pedagogy-when teaching a formal math course-you can't do that. You have to start with axioms and build up from there step-by-step. At each step, you can only use prior results; nothing from the future!