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We know that if a series is convergent, then we can perform grouping on the series and the resulting series would still be convergent.

However,for an arbitrary series, grouping may not always give the same result. E.g. $a_n=(-1)^{n+1}$ is the non-convergent series

$1+(-1)+1+(-1)+...$

which can be grouped to

$(1-1)+(1-1)+...$ which converges to 0.

So my question is, in this link Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ the solutions have grouped the terms in the series (before knowing whether the series converges or not) and showed it is convergent. How is that possible?

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  • $\begingroup$ By grouping the terms and showing that the series whose terms are the respective groups of terms from the original series, it does show that if the original series does happen to be convergent, then the original must converge to the same thing that the grouped series does. That being said, it should still be proven that the original series is convergent as well (but it is unnecessary to find what it converges TO in this step). As for an approach, Dirchlet's Test looks like it should be relevant. $\endgroup$ – JMoravitz May 28 '18 at 4:08
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    $\begingroup$ Side note: $(1-1)+(1-1)+(1-1)+\dots$ converging to zero and $1+(-1+1)+(-1+1)+\dots$ converging to $1$ both simultaneously show that $1-1+1-1+1-1+\dots$ should, if it converges, simultaneously converge to both zero and one, a contradiction. $\endgroup$ – JMoravitz May 28 '18 at 4:09
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Suppose that $s_n=\sum\limits_{k=1}^n a_k$ denotes the $n$-th partial sum. Convergence of the series is equivalent to convergence of the sequence $(s_n)$.

By grouping some term you go to a subsequence $s_{n_k}$. Of course, if $s_{n_k}$ converges, that does not imply in general that so does $s_n$.

However, in the linked post the grouping is made in special way: The answerer took a block of consecutive positive terms, then the next block consists only of negative terms, etc. In this way you get that for $$n_k \le n \le n_{k+1}$$ you have that $s_n$ is between $s_{n_k}$ and $s_{n_{k+1}}$. (Since the sequence of partial sums is monotone on these intervals.)

So in this case if you get that $\lim\limits_{k\to\infty} s_{n_k}=\ell$, you immediately see that the limit of the sequence $(s_n)$ is the same, i.e. $\lim\limits_{n\to\infty} s_n=\ell$.

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  • $\begingroup$ @Anwi The comment "Correct me if I am wrong as $k\to inf$" is somewhat cryptic. I do not know what you are trying to say. $\endgroup$ – Martin Sleziak May 28 '18 at 5:36
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    $\begingroup$ Sorry i am trying to write the latex commands properly $\endgroup$ – Anwi May 28 '18 at 5:38
  • $\begingroup$ Correct me if i am wrong: as k tends to infinity,n tends to infinity(since$ n_k \le n \le n_{k+1}$) so sandwiching we get that lim $s_{n_k} \le $lim $s_n \le$ lim $s_{n_{k+1}}$ $\endgroup$ – Anwi May 28 '18 at 5:44
  • $\begingroup$ @Anwi Well, sandwiching is the basic idea, but you will have $s_{n_k} \le s_n \le s_{n_{k+1}}$ on some intervals and $s_{n_k} \ge s_n \ge s_{n_{k+1}}$ on different intervals. (If you look at odd/even intervals, the sequence $s_n$ is sometimes non-increasing, sometimes non-decreasing.) $\endgroup$ – Martin Sleziak May 28 '18 at 5:48
  • $\begingroup$ True.Other than that,did i miss anything or is that what you meant? $\endgroup$ – Anwi May 28 '18 at 5:49

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