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Show that every sequence in a Banach space such that $\{x_n\} \rightarrow 0$ has a subsequence $\{x_{n_p}\}$ such that $\sum_{p=1}^{\infty} x_{n_p} $ converges by showing $S_N = \sum_{p=1}^{N} x_{n_p}$ is a Cauchy sequence.

The convergent to the zero vector is what is confusing -- typically it is just to some other element? How do I incorporate this?

user79018
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    Why is it confusing? Do you think that (for example) a sequence of real numbers converging to $1$ can have a subsequence whose sum is convergent? – Zev Chonoles May 22 '13 at 21:14
  • Suggestion: Pick your favorite convergent series of positive real numbers, and try to bound the norms of the terms in your subsequence with the terms in the real sequence. – Jonas Meyer May 22 '13 at 21:16
  • @Jonas So if I took 1, 1.4, 1.41, 1.414, 1.4142, ... = ${x_n}$ I can bound them by $\sqrt{2}$ but I don't know how to proceed in the general case. – user79018 May 22 '13 at 21:21
  • I know the sequence converges... Can the subsequence be the whole sequence? – user79018 May 22 '13 at 21:22
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    @user79018: I don't understand your comment. In particular, your sequence has to converge to $0$, so talking about a sequence converging to $\sqrt 2$ isn't relevant. My suggestion was to start by thinking or a convergent series of positive real numbers to use to bound the norms of the terms in your subsequence (the latter being selected carefully such that the bounds hold). Further suggestion: First answer Zev's question above. What is $\lim\limits_{n\to\infty}a_n$ if $\sum\limits_{n\to\infty}a_n$ converges? – Jonas Meyer May 22 '13 at 21:27

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Hint: Since $\lim\limits_{n\to\infty}x_n=0$, then there exist subsequence $\{n_p:p\in\mathbb{N}\}\subset\mathbb{N}$ such that $\Vert x_{n_p}\Vert\leq 2^{-p}$.

Norbert
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