$(a_n)$ is bounded $\implies \sum a_n \frac{1}{n^2}$ converges Show that if $(a_n)$ is a bounded sequence, then
$$\sum \frac{1}{n^2} a_n $$ converges
I only could prove that this sequence is bounded, but not its convergence.
Thanks in advance.

In order to prove convergence it is sufficient to show that the sequence $\{S_M\}_{M\in\mathbb{N}^*}$ given by:
$$ S_M = \sum_{n=1}^{M}\frac{a_n}{n^2}$$
is a Cauchy sequence. Since, assuming $M_1<M_2$:
$$ S_{M_2}-S_{M_1} = \sum_{n=M_1+1}^{M_2}\frac{a_n}{n^2} $$
is bounded in absolute value by:
$$ \sup_{n\in\mathbb{N}^*} |a_n|\cdot\sum_{n=M_1+1}^{M_2}\frac{1}{n^2}\leq \sup_{n\in\mathbb{N}^*} |a_n|\cdot\int_{M_1}^{M_2}\frac{dx}{x^2}=\sup_{n\in\mathbb{N}^*} |a_n|\cdot\left(\frac{1}{M_1}-\frac{1}{M_2}\right)$$
convergence follows.
