Need countereample : If a sequence $(a_n) \in l^2 $ , then the sequence $(1/a_n) \notin l^2 $ I want to know the counterexample for the following statement : Given a sequence $(a_n)$ such that $a_i\ne 0 $ for any $i$ :  If the sequence $(a_n) \in l^2 $ , then the sequence $(1/a_n) \notin l^2$.
If $a\in\ell^2$, then $a_n\to0$. Thus $1/a_n\not\to0$, which prevents it from being in $\ell^2$.
Put $a_n=\frac{1}{n}$.
Then $ \sum \frac{1}{n^2}=\frac{\pi^2}{6}$,finite.

But $\sum n^2$ is not in $l^2$.