$(a_n)_{n=1}^\infty$ 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.

Question

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.

Answer 1

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.

Answer 2

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.