Countable family of Hilbert spaces is complete

Let $H_1, H_2, \ldots, H_n$ be a countable family of Hilbert spaces. Let H be the set of tuples $x = (x_1, \ldots, x_n,\ldots)\in \prod_n H_n$ with the property that $$\|x \| ^2 =\sum_n \| x_n \| _{H_n}^2 <\infty.$$ Show that H is a Hilbert space.

I have no problem showing that that it is a normed, but I have some problem with the completeness. If we define $x^*$ as the coordinatewise limit points. How do we then show that $$\| x^* \| <\infty$$ and $$\lim_{k\rightarrow \infty} \| x^* - x_k \| \rightarrow 0$$

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You use $x_n$ here for both an element of $H_n$ and for a sequence of element in your "new" Hilbert space, which is confusing. – Thomas Andrews Nov 20 '12 at 17:06
Yes, sorry about that, it is changed now. – Johan Nov 20 '12 at 21:02
would this work for showing that $$\|x^*\|$$ is in the set? $$\| x^*\| \leq liminf_{k \rightarrow \infty} \| x^k\| = liminf \| x^k + x^m - x^m\| \leq liminf \| x^k - x^m\| + \|x^m\| \leq \epsilon + b \leq \infty$$. The first inequality is due to Fatou. – Johan Nov 21 '12 at 14:08
"\parallel" produces a rather obscure symbol used for indicating parallel lines. It gives far too much spacing to use as a norm symbol: $\parallel x \parallel$. For norms, use "\|": $\| x \|$. – Chris Eagle Nov 21 '12 at 16:26

Let $(x_n)_{n\geq 1}$ a Cauchy sequence in $H$, then given $\epsilon>0$, there exists $N_o$ positive such that
$$\|x_n-x_m\|_H<\epsilon, \text{ for }n, m\geq N_o\qquad(*).$$ 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 $\Arrowvert\cdot \Arrowvert_H$.$\clubsuit$