So the question says:
If the sequence ($a_n$) (where $n$ $\in$ $\mathbb{N}$) $\rightarrow$ $0$ as $n$ $\rightarrow$ $\infty$, then there exists a subsequence ($a_{n_k}$)(where $k$ $\in$ $\mathbb{N}$) such that
$\sum_{k=1}^{\infty} a_{n_k}$ converges.
Attempt: I found this question vague. I thought it was false, and then I gave a counter-example of
($\frac{1}{n}$), because I thought every sequence is a subsequence of itself and thus the harmonic series diverge.
But on second thought, it seems as if the question is asking : "For any sequence meeting the given condition, there exists a subsequence that satisfies the thing to be proven." Or in other words, maybe there does exist a subsequence of ($\frac{1}{n})$ such that the series converge.
Any help?
Of course it exists. Since $a_n\to0$, there exists $n_1$ such that $|a_{n_1}|\leq1/2$; there exists $n_2>n_1$ with $|a_{n_2}|<1/4$. Continuing in the same fashion, we get $n_1<n_2<\cdots$ with $|a_{n_k}|<2^{-k}$. Then $$ \sum_k|a_{n_k}|\leq\sum_k2^{-k}=1. $$