Let $a_n$ be a sequence such that $\lim\inf |a_n|=0$. How to prove there is a subsequence $(a_{n_k})_{k\in{\mathbb{N}}}$ such that $\sum^\infty_{k=1} a_{n_k}$ converges?
By the definition of cauchy criterion, I think what we want is $\forall\epsilon>0, \exists N$ such that $n\ge m\gt N\implies |\sum^n_{k=m} a_k|<\epsilon$. Now assume $\lim\inf |a_n|=0$, then by deifinition of $\inf$, we know $\forall t<0, \exists N, \forall n>N, |a_n|>t$. Let $\epsilon>0, t=-\epsilon$, then $t<0$, since $|a_n|>t\implies a_n<-t=\epsilon$, we can also say $n>N\implies a_{n}<\frac{\epsilon}{n-m+1}$ where $n\ge m>N$, then we have $|\sum^n_{k=m} a_k|=|a_m+a_{m+1}+...+a_n|<|\frac{(n-m+1)\epsilon}{n-m+1}|=\epsilon$. Since $\sum^\infty_{k=1} a_{n}$ converges, it has subsequence converge to same limit.
However I feel the proof is not rigorous and has errors. Could someone fix it?
Since $\liminf |a_n| = 0$, for every $k\in\mathbb N$, the number $1/2^k$ is not a lower bound of the set $\{a_n\}$. Hence there is an $n_k$ such that $|a_{n_k}| < 1/2^k$. By comparison, $\sum_k |a_{n_k}|\le \sum_k 1/2^k$, so $\sum a_{n_k}$ converges absolutely, and hence in the usual sense.
To make it a little more transparent, we can construct $(a_{n_k})$ inductively as follows. Since $1/2$ is not a lower bound for $(|a_n|)$, there is some $n_1$ such that $|a_{n_1}|<1/2$. Let $n_2>n_1$ be the first index such that $|a_{n_2}|<1/4$, which exists as $\liminf |a_n| = 0$. Let $n_3 > n_2$ be the first index such that $|a_{n_3}| < 1/8$, which exists as $\liminf |a_n| = 0$. Proceed inductively.