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Let $\sum_{k=1}^\infty a_{\varphi(k)}$ be a rearrangement of a conditionally convergent series $\sum_{k=1}^\infty a_k$. Prove that if $\{\varphi(k)-k\}$ is a bounded sequence, then $\sum_{k=1}^\infty a_{\varphi(k)}=\sum_{k=1}^\infty a_k$.

I can't find the solution anywhere and I can't figure it out. Thanks for your help. My understanding of the problem is that if we limit the "space" between the difference of terms then we don't need to reach into an asymptote to find the term $a_{\varphi(k)}$ which creates the same effect as if we were dealing with finite sums instead.

Michael Hardy
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user98964
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  • Can you translate that boundedness condition into a condition on the partial sums? – TBrendle Oct 05 '13 at 17:47
  • My first idea was that for any partial sum for the rearranged series we can find a partial sum of the original series that contains the terms of the rearranged series with the last term being the max. then since we don't need to look at asymptotic behavior to do this we should be able to show the result but I really don't know. I suck at this stuff. – user98964 Oct 05 '13 at 17:52
  • Maybe the partial sums aren't equal, but differ by an amount that vanishes as $n \to \infty$. I'm just guessing. Not sure what you mean by asymptotic behavior. – TBrendle Oct 05 '13 at 17:56
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    I changed {$\varphi(k)-k$} to ${\varphi(k)-k}$, coded as {\varphi(k)-k}. Mixing TeX and non-TeX notation gives results that are ugly and often don't fit together well. – Michael Hardy Oct 05 '13 at 21:18

2 Answers2

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Suppose $$ \sum_{k=1}^\infty a_k = L, $$ and suppose $\phi: \mathbb{N} \to \mathbb{N}$ is such that there is an $N$ for which $\lvert \phi(k)-k \rvert \leq N$ for all $k \in \mathbb{N}$. Let $$S_n = \sum_{k=1}^n a_k$$ and $$S'_n = \sum_{k=1}^n a_{\phi(k)}. $$ Now consider $$ S'_{n+N} - S_n = \sum_{k=1}^{n+N} a_{\phi(k)} - \sum_{k=1}^n a_k. $$ If $n$ is much larger than $N$, then all the terms $a_1$, $a_2$, …, $a_{n}$ appear in $a_{\phi(1)}$, $a_{\phi(2)}$, …, $a_{\phi(n+N)}$, because $N-k \leq \phi(k) \leq N+k$. Let $$A_n=\{\phi(1), \phi(2), \ldots, \phi(n+N) \} \setminus \{1, 2, \ldots, n\} $$ be the set of all indices of terms of $S'_{n+N}$ that don't arise in this way. We have $$A_n \subseteq \{n, n+1, \ldots, n+N\}.$$ Now we have $$S'_{n+N} - S_n = \sum_{k=1}^{n+N} a_{\phi(k)} - \sum_{k=1}^n a_k = \sum_{i \in A_n} a_i,$$ or in other words $$S'_{n+N} = \sum_{i \in A_n} a_i + S_n.$$ By definition, $$\lim_{n \to \infty} S_n = L,$$ while since the series converges $$\lim_{n \to \infty} \sum_{i \in A_n} a_i = 0. $$ Therefore $$\lim_{n \to \infty} S'_{n+N} = L.$$ Thus $$ \sum_{k=1}^\infty a_{\phi(k)} = L, $$ as desired.

TBrendle
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  • This is an awesome solution as well. The use of the tail series to express the notion of our permutated series containing the original series is appealing and I think our professor intended we use it. Thanks! – user98964 Oct 05 '13 at 19:00
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Fix $\varepsilon>0$. Then there exists $k_0$ such that for all $N,M>k_0$ $$\tag{1} \left|\sum_{k=N+1}^M a_k\right|<\varepsilon. $$ By hypothesis, there exists $K$ with $|\varphi(k)-k|<K$ for all $k$. So for all $k=1,\ldots,k_0$, $\varphi(k)\in\{1,\ldots,k_0+K\}$. In other words, the set $\{\varphi(1),\ldots,\varphi(k_0+K)\}$ contains all of $1,\ldots,k$. Then $$ \left|\sum_{k=1}^{k_0+K}a_{\varphi(k)}-a_k\right|\leq\left|\sum_{k:\varphi(k)>k_0}a_{\varphi(k)}\right|<\varepsilon, $$ as the last sum is as in $(1)$.

Martin Argerami
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