Thank you Gigaster for your solution. Here is another solution (similar to Gigaster's solution.)
Obviously, \(\displaystyle{\sum_{k=1}^{1}x_{k}^2=x_{1}^2}\) .
\(\displaystyle{x_1+x_2^2=x_1^2+x_2^2+2\,<x_1,x_2>=x_1^2+x_2^2}\) .
Let \(\displaystyle{\sum_{k=1}^{n}x_{k}^2=\sum_{k=1}^{n}x_{k}^2\,\,(I)}\) for some \(\displaystyle{n\in\mathbb{N}}\) .
Now, by using the fact that the inner product is linear, we get:
\(\displaystyle{\begin{aligned} \sum_{k=1}^{n+1}\,x_{k}^2&=\sum_{k=1}^{n}x_{k}+x_{n+1}^2\\&=\sum_{k=1}^{n}x_{k}^2+2\,<\sum_{k=1}^{n}x_{k}\,x_{n+1}>+x_{n+1}^2\\&\stackrel{(I)}{=}\sum_{k=1}^{n}x_{k}^2+x_{n+1}^2\\&=\sum_{k=1}^{n+1}x_{k}^2\end{aligned}}\)
So, \(\displaystyle{\sum_{k=1}^{n}x_{k}^2=\sum_{k=1}^{n}x_{k}^2\,,\forall\,n\in\mathbb{N}}\) .
If the series \(\displaystyle{\sum_{n=1}^{\infty}x_{n}}\) converges in \(\displaystyle{\left(H,\cdot\right)}\), then, since the \(\displaystyle{\cdot}\)
is continuous, we have that :
\(\displaystyle{\lim_{n\to \infty}\sum_{k=1}^{n}x_{k}^2=\lim_{n\to \infty}\sum_{k=1}^{n}x_{k}^2=\sum_{n=1}^{\infty}x_{n}^2}\) and thus
\(\displaystyle{\sum_{n=1}^{\infty}x_{n}^2=\sum_{n=1}^{\infty}x_{n}^2}\) .
On the other hand, suppose that the series \(\displaystyle{\sum_{n=1}^{\infty}x_{n}^2}\) converges. Since the normedspace \(\displaystyle{\left(H,\cdot\right)}\)
is a \(\displaystyle{\rm{Banach}}\) space, it is sufficient to prove that the sequence \(\displaystyle{\left(\sum_{k=1}^{n}x_{k}\right)_{n\in\mathbb{N}}}\)
is a \(\displaystyle{\rm{Cauchy}}\) sequence. This is true, because, for every \(\displaystyle{n\,,m\in\mathbb{N}\,,n\neq m}\) we get :
\(\displaystyle{\sum_{k=1}^{n}x_{k}\sum_{k=1}^{m}x_{k}\leq \sum_{k=1}^{n}x_{k}+\sum_{k=1}^{m}x_{k}=\sqrt{\sum_{k=1}^{n}x_{k}^2}+\sqrt{\sum_{k=1}^{m}x_{k}^2}}\)
