A question about linear bounded functionals 
Let $X$ be a banach space. $(\phi_n)_n \subset X^*$ a sequence of linear, bounded functionals that satisfies for all $x\in X$, $\sum_n \phi_n(x)$ converges.
Show that $\sum_n \frac{||\phi_n||}{n^2}$ converges.

I am trying to see this using the uniform bounded theorem.
$\sum_n \phi_n(x)$ converges, then the sequence $T_k(\phi_n)=(\sum_{n=1}^{k} \phi_n(x))_{k\geq 1}$ converges so $T_k$ is bounded.And $sup T (\phi_n(x)) < \infty$.Or we can say that there is $N$ such that for every for every $n>N$ $|\phi_n(x)| \leq 1$ so $||\phi_n||=sup {|\phi_n(x)|  , x\in X , \in N} < \infty$ .Thus by the uniform bounded theorem $sup ||\phi_n|| <\infty$.
But how to finish it.

Since for all $x \in X$, the series $\sum_n \phi_n(x)$ converges, we necessarily have $\sup_n |\phi_n(x)| < \infty$ for all $x \in X$. By the uniform boundedness principle, it follows that $M:=\sup_n \|\phi_n \| < \infty$. Consequently
$$\sum_n \frac{\|\phi_n \|}{n^2} \leq M \sum_n \frac{1}{n^2} < \infty$$
as desired.
