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$\sum_{n=1}^{\infty} a_n$ converges to $l$. Let $\{b_n\}$ be any rearrangement of $a_n$'s. We show that $\sum_{n=1}^{\infty} b_n$ also converges.

Let $s_k = \sum_{n=1}^{k} a_n$, and $t_k = \sum_{n=1}^{k} b_n$ be partial sums. Since for any $\epsilon > 0$, there exists a number $N$ such that $\forall n \geq N$, $|s_n - l| < \epsilon$. Or, equivalently $|s_m - s_N| < \epsilon$ for any $m>N$. (also called Cauchy criterion)

Now, in the rearrangement series, choose an $M$ such that $a_1, a_2, \dots, a_N$ are covered in $b_1, b_2, \dots, b_M$. Therefore $|t_M - s_N| = |a_{i_1} + a_{i_2} + \dots + a_{i_{M-N}}| \leq |s_{M'} - s_N|$ for some $M' > N$. But since $|s_m - s_N| < \epsilon$ for any $m>N$, we have $|t_M - s_N| < \epsilon$. Consequently, $|t_M - l| < 2\epsilon$. And the convergence of rearrangement follows.

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    $\begingroup$ You cannot conclude $\ldots \le |s_{M'}-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms $\endgroup$ – Hagen von Eitzen Jul 22 '18 at 17:05
  • $\begingroup$ Gotcha! Thanks. $\endgroup$ – student Jul 22 '18 at 18:51
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Hard to say exactly what's wrong because it's hard to know exactly what you had in mind - in particular it seems to me that the $|a_{i_1}+\dots|$ should be $|b_{i_1}+\dots|$. But if I look at what you wrote and try to make a correct proof out of it, here's a hint regarding where the error is: If $a_1=1$, $a_2=-1$ and $a_3=1$ then $$|a_1+a_3|>|a_1+a_2+a_3|.$$

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As Hagen von Eitzen also pointed out, the problem lies in taking the following inequality:

$$ |a_{i_1} + a_{i_2} + ... + a_{i_{M-N}}| \leq |s_{M'} - s_N| = |a_{N+1} + a_{N+2} + ... + a_{M'}| $$

The indices $i_1, ..., i_{M-N}$ are between $N+1$ and $M'$, and so the inequality would hold if all the $a_k$ were positive. However, this is not the case, and so we cannot take the inequality. E.g., we have:

$$ |1+1| \nleq |1+(-1)+1| $$

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  • $\begingroup$ Thanks @Sambo . $\endgroup$ – student Jul 22 '18 at 18:51

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