Summing The Alternating Harmonic Series

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[HershO.]

While simplifying f'(x), we supposed that -1<x<1. Then how are we allowed to plug in x =1? analytic continuation maybe?

[alextang]

4:36 In this step, you don't need to separate the terms anymore because the ratio(r) of GS is -x. So 😋😋😋😋😋😋

[scottleung]

Wait a minute, are you now publishing two regular-length vids per day?

[tanvirasif]

5:54 instead of breaking it into 2 parts we can take common ratio as '-x' and that will make things easy

[ahmetkaraagac]

Let S=Sum of infinite 1/n (1, n) series
K=S-0.5S= 0.5S (Terms with even denominator eliminated->1/2, 1/4, ...) We obtain->1+1/3+1/5+....
K-0.5S=0 (Negative of even denominator terms added, so we obtain the terms in your question. Which turns out to be ZERO.

Simply without derivative, integral, limit etc.

[vladimirrodriguez]

Great and smart resolution. We want more please

[slr]

I understand how this works after watching this video, but could you explain the thought process behind selecting a function and differentiating it to solve this?

[mismis]

What I did when faced with this problem is try to find the taylor series, and find a function whose Nth derivative is proportionnal to (n-1)!. Luckily, the log function satisfies this, you have to then center it at one, and so you get the expected result.
I also tried doing it the way you did it but was not able to, so thanks a lot for that !

[erikroberts]

WolframAlpha should have told you that this series converges conditionally, instead of just telling you that it converges. Just like a divergent series, you should always avoid conditional convergencies whenever possible in mathematic equations.

[morteza]

First, we edit the question in this way;
Σ(-1)^(n-1)(1/n) n≥1 n∈N
Then we use the following formula which is actually a definite integral;
∫[0, 1]x^(n-1)dx=1/n
So we put it in the question;
Σ(-1)^(n-1)∫[0, 1]x^(n-1)dx n≥1 n∈N
We know that our series is convergent
So ( Σ∫=∫Σ )
Then we use this technique;
∫[0, 1](Σ(-1x)^(n-1))dx n≥1 n∈N
This is a geometric series, so we treat it according to the rules of geometric series ;
∫[0, 1](1/(1-(-x))dx=ln(1+1)-ln(1+0)=ln(2)

[milandus]

It seems to me, that the original problem deals only with the "coefficients" (i.e., number 1 and then the fractions).
So then, why is it needed, in defining the f(x), to moreover work with the powers of "x" ???

[alexl]

So for x = 2 we would have ln(3) = 2 - 2 + 8/3 - 4 + 32/5 - 64/6 + ... Is it corerct? Looks like such series does not converge

[rotten-Z]

This relation for series is valid for x≤1 only? How its possible when infinite sum of powers of 2 (for example) is equal to 1/(1-2)?

[abbasfarazdel]

This discussion is entirely incorrect because 1+r+r2+… = 1/(1-r) is valid ONLY when |r|<1.
It is well known that all conditionally convergent series can be equal to any arbitrary value by reordering the infinite series.

[paulokas]

6:00 so by using that formula one finds the f'(x) just for x in ]-1, 1[
But the goal was to find f(1).
So what's the point of finding the derivative with no value in x=1

[m]

Whew! Half of what he said went over my head. The other half went way over…

[ziplock]

Wow
Amazing... so, how is pi/4 = 1 - 1/3 + 1/5 ...?

[rsactuary]

don't you have to prove that the derivative converges before you start moving terms around?

[Qermaq]

My method is more fun if less accurate. Let 1 + 1/2 + 1/3 + 1/4 + .. = S. If I subtract S from itself sneakily I get S-S = 0 = 1 + 1/2 (-1) + 1/3 + 1/4 (-1/2) +1/5 + 1/6 (-1/3) + ... = 1 -1/2 + 1/3 - /14 + ....

I know. But isn't it fun?

[louisromao]

To answer your early question:
1 + 1/2 + 1/3 + 1/4 + 1/5 + ... is the harmonic series and it diverges.