Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequance of pairwise orthogonal vectors in a Hilbert space $H$. Show that the following are equavalent:
(a) $\sum_{n=0}^\infty x_n$ converges in the norm topology of $H$.
(b) $\sum_{n=0}^\infty \|x_n\|^2 < \infty$.
(c) $\sum_{n=0}^\infty \langle x_n, y \rangle$ converges for every $y\in H$.
I can prove that (a) implies (b), but none of the other implications.
$(a)\iff (b)$: You can write $x_{n}=\alpha_{n}e_{n}$ where $\{ e_{n}\}_{n=1}^{\infty}$ is an orthonormal subset of $X$. Then $\sum_{n}x_{n}=\sum_{n}\alpha_{n}e_{n}$ converges iff $\sum_{n}|\alpha_{n}|^{2} = \sum_{n}\|x_{n}\|^{2} < \infty$.
$(a)\implies (c)$: If $\sum_{n}x_{n}$ converges, then $(\sum_{n}x_{n},y)=\sum_{n}(x_{n},y)$ converges, regardless of $y$, by continuity of the inner-product with respect to norm convergence.
This part (c)$\implies$ (b) is the trickiest, and I had that wrong, as pointed out by Jonas Meyer. Following the comment of PhoemueX above, suppose $\sum_{n}(x_{n},y)$ converges for all $y$, and define linear functionals $F_{k}(y)=\sum_{n=1}^{k}(y,x_{n})$. Each $F_{k}$ is bounded and $\sup_{k}|F_{k}(y)| < \infty$ for all $y$. Therefore, $\|F_{k}\| \le M$ for some constant $M$ by the Uniform Boundedness Principle. In particular, for $k \ge l$, $$ \sum_{n=1}^{l}\|x_{n}\|^{2}=|F_{k}(\sum_{n=1}^{l}x_{n})| \le M\|\sum_{n=1}^{l}x_{n}\| = M\left(\sum_{n=1}^{k}\|x_{n}\|^{2}\right)^{1/2}, $$ which implies the following for all $l$: $$ \left(\sum_{n=1}^{l}\|x_{n}\|^{2}\right)^{1/2} \le M. $$ So $(c)\implies (b)$ per PhoemueX's suggestion.