Prove that $\sum_{n=0}^{\infty} x_n $ convergent in Hilbert space

Question

Let $(x_n)_{n\in \mathbb{N}}$ be a sequence in Hilbert space such that $$ \displaystyle \sum_{n=0}^{\infty} ||x_n||^2 < +\infty$$ Assume that there exists an integer $N_0$ such that, if $|n-m| \geq N_0$, the vector $x_n $ and $x_m$ are orthogonal. Prove that the series $ \displaystyle \sum_{n=0}^{\infty} x_n$ is convergent and there exist a constant C that only depends on $N_0$, such that $$ ||\displaystyle \sum_{n=0}^{\infty} x_n||^2 \leq C \displaystyle \sum_{n=0}^{\infty} ||x_n||^2$$ My attempt: For any $\epsilon > 0$ there exists an integer $N_1$ such that $$\displaystyle \sum_{n=N_1+1}^{\infty} ||x_n||^2 < \epsilon$$
Take $N = \max\{N_0; N_1\}$ and if let $S_p = \displaystyle \sum_{n=0}^{p} x_n$, for all $p > 0$, we have:
$$ || S_{N+p}-S_N||^2 = ||\displaystyle \sum_{n=N+1}^{N+p} x_n||^2 = \displaystyle \sum_{n=N+1}^{N+p} ||x_n||^2 + 2\displaystyle \sum_{m>n>N} (x_m|x_n) $$ where (.|.) is the inner product of the Hilbert space. But from here, I don't know how to use the assumption to prove the left-hand side to be less than $\epsilon$. Please help me prove and find the constant C.

Answer 1

\begin{align*}\|\sum_{n=0}^Nx_n\|^2&\le\sum_{n=0}^N\|x_n\|^2+2\sum_{n,m=0}^N|(x_n|x_m)|\\ &=\sum_{n=0}^N\|x_n\|^2+2\sum_{n=0}^N\sum_{m=n-N_0}^{n+N_0}|(x_n|x_m)|\\ &\le\sum_{n=0}^N\|x_n\|^2+\sum_{n=0}^N\sum_{m=n-N_0}^{n+N_0}(\|x_n\|^2+\|x_m\|^2)\\ &\le(2N_0+2)\sum_{n=0}^N\|x_n\|^2+\sum_{i=-N_0}^{N_0}\sum_{n=0}^N\|x_{n+i}\|^2\\ &\qquad\textrm{The last sum is relabeled along diagonals on an $n$-$m$ array.}\\ &\qquad\textrm{Take $x_{n+i}=0$ for $n+i<0$.}\\ &\le2(2N_0+2)\sum_{n=0}^{N+N_0}\|x_n\|^2 \end{align*} The required inequality follows from $N\to\infty$.

Convergence of $\sum_nx_n$ now follows by showing it is Cauchy, by extending this argument to $\|\sum_{n=N}^Mx_n\|^2\le C\sum_{n=N}^M\|x_n\|^2$.