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Riemann's theorem states that if a series is conditionally convergent, then for any number $L$ (could be infinite), the series can be rearranged in such manner that it would converge to $L$. I was wondering, is the converse true?

More formally, let $a_n$ be a sequence such that $a_n\to 0$, and for every $L$ (could be infinite), there exists a permutation $\sigma$ such that $\sum\limits_{n=1}^{\infty}a_{\sigma(n)} = L$. Does this necessarily mean that $\sum\limits_{n=1}^{\infty} a_n$ converges to a finite number? (it must be conditional convergence)

Joshhh
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2 Answers2

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The converse does not hold. For start with the usual series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$.

This can be rearranged so that the rearranged series does not converge. Let $a_1+a_2+a_3+\cdots$ be such a rearrangement. Then $a_1+a_2+a_3+\cdots$ does not converge, but the terms can be rearranged to give any desired sum.

André Nicolas
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By your definition of $a_n$, there is a permutation $σ$ such that $∑_{n=1}^∞a_{σ(n)}=∞$. Now take $b_n=a_{σ(n)}$, then $b_n$ satisfies your conditions (as permutations form a group) and $∑_{n=1}^∞b_n=∞$.

jugglingmike
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    Nice way to do it...............After-thought: But what if we do not state that some permutation could sum to infinity? – DanielWainfleet Jun 25 '16 at 18:38
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    @user254665: Same result. We know that the series is not absolutely convergent, since all permutations of an absolutely convergent real series converge to the same value. So any convergent permutation must be a conditionally convergent permutation, so Riemann's theorem applies, thereby guaranteeing the existence of a permutation that diverges to ∞. – ruakh Jun 25 '16 at 20:27