# $(a_n)_{n=1}^\infty$ s.t $\lim_{n \rightarrow \infty} |a_n| = 0$. Prove $\exists$ subsequence $(a_{n_k})$ s.t $\sum_{k=1}^\infty a_{n_k}$ converges.

Let $$(a_n)_{n=1}^\infty$$ be a sequence such that $$\lim_{n \rightarrow \infty} |a_n| = 0$$. Prove that there is a subsequence $$(a_{n_k})$$ such that $$\sum_{k=1}^\infty a_{n_k}$$ converges.

Stuck on this question. I know that if we have a sequence $$x_n$$ that converges $$\rightarrow 0$$, $$\sum x_n$$ converges but the converse isn't necessarily true. The absolute value had me consider using the Comparison Test.

My initial thoughts with this is using that the sequence is convergent and there bounded. Then by Bolzano-Weirstrass, we know there exists a convergent subsequence. And then using some fact to show that the subsequence converges.

But not too sure, would appreciate help.

## 2 Answers

Since $$a_n \to 0$$, for each $$k\in \mathbb{N}$$, there exists an integer $$n_k \in \mathbb{N}$$ such that $$|a_{n_k}| \leq 2^{-k}$$. It follows that $$\sum\limits_{k=1}^\infty |a_{n_k}|$$ converges as it is bounded above by a converging series.

• Is $2^{-k}$ just an arbitrary choice to bound it since we know that the sequence converges to 0, so that inequality is true eventually? – SS' Oct 25 '18 at 18:22
• yes, it's an arbitrary choice of a converging series. You can take anything you like which converges, say $1/k^2$ for instance. Then fixing the $k$-th term you can make $|a_n|$ smaller than that by choosing $n$ large enough thanks to the condition that $a_n \to 0$. – Hayk Oct 25 '18 at 18:24

Let $$f(1)$$ be the least $$j$$ such that $$|a_j|<1/2.$$ For each $$n\in \Bbb N$$ let $$f(n+1)$$ be the least $$j$$ such that (I) $$j>f(n)$$ and (II) $$|a_j|<2^{-(n+1)}.$$

Then $$(f(n))_{n\in \Bbb N}$$ is a strictly increasing sequence, and $$\sum_{n=1}^{\infty}|a_{f(n)}|<\sum_{n=1}^{\infty}2^{-n}=1<\infty.$$

So $$\sum_{n=1}^{\infty}a_{f(n)}$$ converges.