$X$ is a Banach space and $(\phi_n)_n\subset X^*$ is a sequence such that for all $x\in X$ the series $\sum\limits_n\phi_n(x)$ converges. Show that $\sum\limits_n\frac{\|\phi_n\|}{2^n}$ converges.
The hypothesis seems to suggest that we define $f_m(x)=\sum\limits_{n=1}^m\phi_n(x)$. Obviously each $f_m$ is linear and continuous. Since for each $x$ the above infinite series converges, the family $(f_m)_m$ is pointwise bounded. Thus also uniformly bounded. However this does not lead me to the solution. Any hint would be appreciated.
So $\phi_{n}(x)\rightarrow 0$ by the summability, and hence $(\phi_{n}(x))_{n=1}^{\infty}$ is a bounded set of scalars, then by UBP, $\sup\|\phi_{n}\|:=M<\infty$ and hence $\dfrac{\|\phi_{n}\|}{2^{n}}\leq\dfrac{M}{2^{n}}$.