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I am given a question: Can a series with non-negative terms be turned via rearrangement into series with different sum? My solution: No:

If $\sum a_n$ is a series with non-negative terms, then the sequence of partial sums given by $s_n=\sum_{k=1}^n$ is monotone increasing, thus has a limit in $\mathbb{R} \cup \{ \infty \}$. So, if $\sum a_n$ converges, then $\sum|a_n|$ converges too, because $a_n\geq 0$ as assumed. From Riemann rearangement theorem any of its reordering converges and has the same sum. I am stuck with the part where $\sum a_n$ diverges to $\infty$. I wish to show that for any bijection $\varphi:\mathbb{N}\rightarrow \mathbb{N}$ the series $\sum a_{\varphi(n)}$ diverges to $\infty$. I tried looking at $$ s'_n=\sum_{k=0}^n a_{\varphi(k)} $$ and show that this diverges. Isn't this somehow related to the sequence or the subsequence of $s_n$? How to proceed?

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Let $\sum_{m=1}^\infty a_{\phi(m)}\le K$ for some $K>0$. Consider a partial sum of $\sum_{n=1}^Na_n$. Then for $n=1,\ldots,N$ there exists $m_n$ such that $a_n = a_{\phi(m_n)}$. Hence, setting $M = \max\{m_n : n=1,\ldots,N\}$, $$ s_N = \sum_{n=1}^Na_n = \sum_{n=1}^Na_{\phi(m_n)}\le\sum_{m=1}^{M}a_{\phi(m)}\le\sum_{m=1}^\infty a_{\phi(m)}\le K. $$ This holds for every $N$ and hence $(s_N)$ is increasing and bounded, thus convergent.

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    $\begingroup$ Just got to the very same result (idea), wonderful. Thanks! $\endgroup$ – Michal Dvořák Nov 1 at 21:15

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