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I'm testing ideas about the convergence and divergence of series:

Say I have a series $$ 1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + . . .$$

I believe that this series would converge to $0$ as every odd term in the series is canceled by the following term. (In addition, the individual terms are decreasing to 0 from 1 and -1.) In other words, the sequence of partial sums of this series tend to a limit, $0$.

Say I have a different series $$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{2^3} + \frac{1}{4} - \frac{1}{2^4} + . . .$$

My assumption would be that the sum of odd terms is not canceled out by the sum of even terms. In other words, this series would increase to infinity. The partial sum of the first $i$ terms would be less than the partial sum of the first $i+2$ terms, and so forth.

How do these series converge or diverge? Am I looking at this the right way?

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    $\begingroup$ Perhaps worth explicitly noting in your first example: Convergence occurs because the individual terms are going to $0$. For instance the series $1 - 1 + 2 - 2 + 3 - 3 + ...$ diverges, even though it also has the property that every other partial sum is $0$. $\endgroup$
    – paw88789
    Nov 3, 2021 at 21:49
  • $\begingroup$ Yes, very important to state definitely. Were it the other way around, as you used in your example, my thought would be incorrect despite the similarity. $\endgroup$ Nov 3, 2021 at 21:51
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    $\begingroup$ You are right that the first series converges and the second diverges to $+\infty$. To prove these facts: for the first one you can see that the even partial sums are $0$ while the odd partial sums are $1, 1/2, 1/3, \ldots$; for the second observe that the partial sums are the differences of partial sums of the divergent series $1 + 1/2 + 1 + 1/4 + \ldots$ and the convergent series $1/2 + 1/4 + 1/8 + \ldots$. $\endgroup$
    – Rob Arthan
    Nov 3, 2021 at 21:54
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    $\begingroup$ How @RobArthan's points look formally: let $a_n$ ($b_n$) denote the sum of the first $n$ terms in your first (second) series. Since $a_{2n}=0$ and $a_{2n-1}=\frac1n$ for $n\ge1$, the first series converges to $0$. And since $b_{2n}=\underbrace{H_n}_{\sim\ln n}-\underbrace{\sum_{k=1}^n2^{-k}}_{\sim1}$, the second series diverges to $\infty$. $\endgroup$
    – J.G.
    Nov 3, 2021 at 21:55
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    $\begingroup$ Does anyone have thoughts about conditional vs absolute convergence for the first of these series? My thoughts were that taking the absolute value of the terms would make it diverge to infinity, meaning it would be conditionally convergent. $\endgroup$ Nov 3, 2021 at 23:03

2 Answers 2

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$$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{2^3} + \frac{1}{4} - \frac{1}{2^4} + . . .$$

can be coded as $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big)$. If this series would converge, as the sum of convergent series is a convergent series, then $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big) +\sum_{n\geq1}\frac{1}{2^n}=\sum_{n\geq1}\frac{1}{n}$ would converge. This is a absurd, therefore the series diverges.

The reasoning behind your conclusion - "The partial sum of the first $i$ terms would be less than the partial sum of the first $i+2$ terms, and so forth" might be wrong... take for example $e=\sum_{n\geq0}\frac{1}{n!}$.

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Your reasoning for the first one is correct. For the second one, your argument doesn't rule out the possibility that the positive terms alone form a convergent sum. But indeed they are a well-known divergent series whereas the sum of the negative terms converges. From this we can deduce that the whole thing diverges.

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