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"Prove that if $a_n \to 0$, then there is a sub-sequence $a_{n_k}$ such that $\sum_{k} a_{n_k}$ converges."

I don't even know how to start. Can anyone help?

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Take a subsequence $a_{n_k}$ for which $\lvert a_{n_k} \rvert < 2^{-k}$. Then you are done.

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Since $a_n \to 0$, we have that for every $\epsilon$ there is $N_{\epsilon}$ such that for all $n \geq N$, we have $|a_n| < \epsilon$.

In particular, for every $\epsilon = \frac{1}{m^2}$ there is $N_\epsilon$ such that $|a_{N_\epsilon}| < \frac{1}{m^2}$.

So pick out the sequence $a_{N_m}$. We know that $\sum a_{N_m}$ converges absolutely by comparison with the convergent $\sum_{i=1}^{\infty} \frac{1}{i^2} = \frac{\pi^2}{6}$.

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    To be rigorous, you have to pick $N_m$ to be strictly increasing. Otherwise $a_{N_m}$ is not a subsequence. – 2017-01-21
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    By that logic, doesn't it mean that $a_{n}$ also converges? If $a_{n}$ $\to$ 0, it doesn't mean that $\sum$ $a_{n}$ converges (ex. $a_{n}$=1/n) – 2017-01-21
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    @TonyK Yes, fair point. (RonLibman: the reason the sum converges is by comparison with a known-convergent series.) – 2017-01-21