I'm trying to solve this problem,
Let $a_n$ be a sequence such that lim inf$ |a_n| = 0 $. Prove that there is a subsequence $a_{n_k}$ such that $\sum a_{n_k}$ converges.
So far, I tried to say we can deduce that $a_n > K$ where $0 < K < \infty$ and $n > N$ for some $ N $. So $\sum a_n > K > 0$. After this I do not know how I can finish this proof. Any advice/solutions would be appreciated.
$\liminf |a_n|=0$ means that $\lim_{n \to \infty} \big( \inf_{m \geq n} |a_m| \big)= 0$.
This means that for any $\epsilon > 0$, you can find $N$ such that $\inf \{|a_m|:m \geq n \} < \epsilon$ for all $n \geq N$. This in turn implies that for any $\epsilon>0$ there are arbitrarily large $n$ such that $|a_n|< \epsilon.$ (prove this).
We will use this fact in order to inductively construct a sequence $n_k$ of increasing indices such that $a_{n_k} < 2^{-k}$ for all $k$. Choose $n_1 \in \mathbb{N}$ such that $|a_{n_1}|< \frac{1}{2}$. For the inductive step, suppose we have constructed $n_k$. Then let $\epsilon=2^{-(k+1)}$. By the bold statement above, we can choose $n_{k+1} \in \mathbb{N}$ such that $n_{k+1}>n_k$ and $a_{n_{k+1}}<\epsilon$. Continuing like this inductively, we get our desired sequence $n_k$ of indices.
Since $a_{n_k}< 2^{-k}$ for all $k$, it follows that $\sum_k a_{n_k}$ converges.