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How would I go about proving that if $a_n$ is a real sequence such that $\lim_{n\to\infty}|a_n|=0$, then there exists a subsequence of $a_n$, which we call $a_{n_k}$, such that $\sum_{k=1}^\infty a_{n_k}$ is convergent.

I think that I can choose terms $a_{n_k}$ such that they are terms of a geometric series, so that means that it will converge, but I don't know how to formally state this.

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2 Answers 2

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Your idea is good. You can pick $a_{k_1}$ such that $|a_{k_1}|<1/2$. Then pick $k_2>k_1$ so that $|a_{k_2}|<1/4$, and inductively pick ${k_n}>{k_{n-1}}$ such that $|a_{k_n}|<1/2^n$.

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  • $\begingroup$ Should it be $a_{k_n}<1/2^{n-1}$? $\endgroup$ Mar 13, 2015 at 2:14
  • $\begingroup$ I corrected it! $\endgroup$
    – Dimitris
    Mar 13, 2015 at 2:14
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Start with a series you know is convergent. A geometric series will work as you have guessed. I'll use the series $\sum_{n=1}^\infty \frac{1}{n^2}$ for my example. Choose each $a_{n_{k}}$ such that $\left|a_{n_{k}}\right|\leq \frac{1}{n_k^2}$ for some integer $n_k$. You will need to prove that you can find infinitely many of these $a_{n_{k}}$. Once you know that, then $$\sum_{n=1}^\infty \left|a_{n_{k}}\right|\leq \sum_{k=1}^\infty \frac{1}{n_k^2} \leq \sum_{n=1}^\infty \frac{1}{n^2}$$ Hence, $$\sum_{n=1}^\infty a_{n_{k}}$$ is an absolutely convergent series.

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