Let $(x_n)_{n\geq 1}$ a Cauchy sequence in $H$, then given $\epsilon>0$, there exists $N_o$ positive such that
\begin{equation} \|x_n-x_m\|_H<\epsilon, \text{ for }n, m\geq N_o\qquad(*). \end{equation} from this equation we have, for each $k\geq 1$
$\|x_n^k-x_m^k\|_{H_k}^2\leq\sum_{k}\|x_n^k-x_m^k\|_{H_k}^2=\|x_n-x_m\|^2<\epsilon^2;\text{ for } n, m\geq N_o.$ This means that $(x_n^k)_{n\geq 1}$ is a cauchy secuence in $H_k$ for each $k\geq 1,$ so there exists $x_k\in H_k$ such that $x_n^k\to x_k$ as $n\to \infty.$ Now we will consider the element $x=(x_1,x_2,x_3,x_4,\ldots,x_n, \ldots )$ obtained in this way. It's not too much dificult to show that $x\in H$ and $x_n\to x$ as $n\to \infty.$ In order to show that $x\in H$, we will show that $x_{n_o}-x\in H$ for some $n_o,$ since $H$ is closed under subtraction it will follow that $x=x_{n_o}-(x_{n_o}-x)\in H.$ Pick $N\in \mathbb{N}$, by $(*)$ $\sum_{k=1}^{N}\Arrowvert x_n^k-x_m^k\Arrowvert_{H_k}^2\leq \sum_{k}\Arrowvert x_n^k-x_m^k\Arrowvert_{H_k}^2=\Arrowvert x_n-x_m\Arrowvert_H^2<\epsilon^2, \text{ for } n,m\geq N_o.$ Let $m\to \infty$ in the finite sum on the left hand side: we obtain $\sum_{k=1}^{N}\Arrowvert x_n^k-x_k\Arrowvert_{H_k}^2<\epsilon^2, \text{ for } n\geq N_o,$ Since this equation holds for all $N\in \mathbb{N}$ we have, on letting $N\to \infty$, $\sum_{k=1}^{\infty}\Arrowvert x_n^k-x_k\Arrowvert_{H_k}^2<\epsilon^2, \text{ for } n\geq N_o,$ that is $\Arrowvert x_n-x\Arrowvert_H<\epsilon, \text{ for } n\geq N_o, \qquad(**)$ In particular we have $\Arrowvert x_{N_o}-x\Arrowvert_H<\epsilon,$ and so $x_{N_o}-x\in H, $ this proves that $x\in H,$ at the same time from $(**)$ we get $x_n\to x$ as $n\to\infty$ with recpect the norm $\Arrowvert\cdot \Arrowvert_H$.$\clubsuit$