Proving a series of a subsequence converges

Solution 1:

If $a_{n}$ is unbounded, it means that $\forall C>0$, it exists $\bar n\in\mathbb N$ such that $|a_{\bar n}|>C$. In particular, it is easy to prove that if $(a_{n})_{n>0}$ is unbounded, then $(a_{n})_{n>k}$ is unbounded, too, $\forall k\in\mathbb N$.

Consider a sequence $C_{k}$ such that $\sum 1/C_{k}<+\infty$ (for example $C_{k}=k^2$). Define $$n_{k}:=\min\{n>n_{k-1}\ \text{such that}\ |a_{n}|>C_{k}\},$$ which is well defined by the remarks done above. It follows that $$\sum_{k=1}^{\infty}\frac{1}{|a_{n_{k}}|}\le\sum_{k=1}^{\infty}\frac{1}{C_{k}}<+\infty,$$ that is series converges absolutely.