Need countereample : If a sequence $(a_n) \in l^2 $ , then the sequence $(1/a_n) \notin l^2 $

Question

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 $(1/a_n) \notin l^2$.

Answer 1

Note that $$\left(\sum_{\nu =1}^n 1\right)^2=n^2\leqslant \sum_{\nu=1}^n a_\nu^2\sum_{\nu =1}^n a_\nu^{-2}$$

This gives a pretty rough estimate of how the sum of the squares of the reciprocals diverge. Of course the simplest is Martin's approach.

Answer 2

If $a\in\ell^2$, then $a_n\to0$. Thus $1/a_n\not\to0$, which prevents it from being in $\ell^2$.

Answer 3

Put $a_n=\frac{1}{n}$. Then $ \sum \frac{1}{n^2}=\frac{\pi^2}{6}$,finite.

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