There is an implicit question about the convergence of the sums, which this answer does not address.
If $\sum_{n=1}^\infty x_n$ converges in the norm of the Hilbert space $H$, then for any $y \in H$, $\sum_{n=1}^\infty \langle x_n, y \rangle$ converges in the base field ($\mathbb{R}$ or $\mathbb{C}$), and $\langle \sum_{n=1}^\infty x_n , y \rangle = \sum_{n=1}^\infty \langle x_n, y \rangle.$ As mentioned, this is because the inner product is linear and continuous with respect to the $H$ norm topology (essentially, by Cauchy-Schwarz).
However, if the sum on the right side converges for some $y \in H$, or even for all $y \in H$, it does not follow that $\sum_{n=1}^\infty x_n$ converges in $H$, so the left side may be meaningless. For an example, let $\{e_i\}_{i=1}^\infty$ be an orthonormal set in $H$ (assuming it is infinite dimensional). Let $x_1 = e_1$, and $x_n = e_n - e_{n-1}$ for $n \ge 2$. Then $\sum_{n=1}^\infty x_n = \lim_{i \to \infty} e_i$ does not converge in norm. However, $\sum_{n=1}^\infty \langle x_n, y \rangle = \lim_{i \to \infty} \langle e_i, y \rangle = 0$ for all $y$, by Bessel's inequality.
Effectively, having the right side converge for all $y$ only implies that $\sum_{n=1}^\infty x_n$ converges in the weak topology.