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Question:

A subseries of the series $\sum _{n=1}^\infty a_n$ is defined to be a series of the form $\sum _{n=1}^\infty a_{n_k}$, for $n_k \subseteq \Bbb N$. Prove that $\sum _{n=1}^\infty a_n$ converges absolutely if and only if each subseries $\sum _{n=1}^\infty a_{n_k}$ converges.

Suggested solution:

$\Rightarrow$

  • We assume $\sum _{n=1}^\infty a_n$ converges absolutely $\Rightarrow \lim_{n \to \infty} |a_n|$=0. We know from the definition of series that it's actually the sequence of partial sums so $\sum _{n=1}^\infty a_n = S_n$. Therefore we can treat it like a sequence and say that it converges to L. Since $S_n$ converges to L, each of it's sub sequences also converge to L. Therefore $\sum _{n=1}^\infty a_{n_k} =S_{n_k} \to L$ as well.

$\Leftarrow$

  • We assume each subseries converges. Specifically the sub series $\sum _{\Bbb N even} a_n = A$ $\sum _{\Bbb N odd} a_n = B$ . Since these two series comprise all of the naturals, then $\sum _{n=1}^\infty a_n =\sum _{\Bbb N even} a_n+\sum _{\Bbb N odd} a_n = A+B$ . Therefore it converges to A+B.

I would like to verify this proof, because I've given it a lot of thought, and still not 100 percent sure about it. Please hint me, and notify me about any mistakes. Thanks.

jreing
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  • In $\implies$: (1) $\sum_{n=1}^\infty a_n = S_n$ makes no sense; you probably want $\sum_{n=1}^\infty a_n = \lim_{n\rightarrow \infty} S_n$. (2) It isn't true that $S_{n_k} = \sum_{n=1}^\infty a_{n_k}$. You are confusing a subsequence of $a_n$ with a subsequence of $S_n$. For instance, take $a_n = 0$ if $n$ is odd, and $a_n = 1/2^n$ if $n$ is even. Then $\sum_{n = 1}^\infty a_{2n + 1} = 0$ but $\lim_{n \rightarrow \infty } S_{2n+1} = \lim_{n \rightarrow \infty}\sum_{m=1}^{2n+1} a_m= \sum_{m=1}^\infty a_m > 1/4 \neq 0$. –  May 07 '13 at 19:55
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    In the other implication, if each subseries converges absolutely, then in particular $\sum_{n=1}^\infty a_n$ converges absolutely (since $a_n$ is a subsequence of $a_n$). –  May 07 '13 at 19:56
  • @111: Not relevant to this question, but it's amazing to me how many people think ${a_n}$ isn't a subsequence of ${a_n}$. – wj32 May 07 '13 at 21:23

1 Answers1

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The two suggested proofs are wrong

$\Rightarrow$

We have $$\sum_{k=1}^\infty|a_{n_k}|=0\times|a_1|+\cdots+0\times|a_{n_1-1}|+|a_{n_1}|+0\times|a_{n_1+1}|+\cdots\leq\sum_{n=1}^\infty|a_{n}|$$ hence the series $$\sum_{k=1}^\infty|a_{n_k}|$$ is convergent by comparaison.

$\Leftarrow$

Let denote by $(a_{n_k})$ the subsequence of $(a_n)$ by choosing all its positive terms and $(a_{n'_k})$ the subsequence of $(a_n)$ by choosing all its negative terms then the series $$\sum_{k=1}^\infty a_{n_k} \quad\text{and} \quad \sum_{k=1}^\infty a_{n'_k}$$ are absolutely convergent and its clear that $$\sum_{n=1}^\infty |a_{n}|=\sum_{k=1}^\infty a_{n_k}-\sum_{k=1}^\infty a_{n'_k}$$ hence the series $$\sum_{n=1}^\infty a_{n}$$ is absolutely convergent.