Cauchy sequence $S_{n}= \sum_{j=1}^{n} \langle x,x_{\alpha}\rangle x_{a}$ in Hilbert space

Question

Let $(x_{\alpha})_{\alpha \in I}$ be an orthonormal family of vectors in a Hilbert space H and let $J:=${$\alpha$ $\in$$I$ $|$ $\langle$$x$,$x_{\alpha}$$\rangle$$\neq$$0$} with {$\alpha_{1}$$,$$\alpha_{2}$$,$$...$} an enumeration of $J$ (it is shown elsewhere that $J$ is countable). Fix $x\in H$. For the theorem I'm looking at I want to show that when {$\alpha_{1}$$,$$\alpha_{2}$$,$$...$} is an enumeration of $J$ and $$S_{n}= \sum_{j=1}^{n} \langle x,x_{\alpha_{j}}\rangle x_{\alpha_{j}}$$ that $(S_{n})_{n}$ is convergent. I understand that since this sequence is in a Hilbert space that we just need to show that it is Cauchy. I have that for $n\leq$ $m$ $\left\lVert S_{n}-S_{m}\right\rVert$$^{2}$$=$$\sum_{j=m+1}^{n}|\langle x,x_{\alpha_{j}}\rangle|$$^{2}$. The proof I'm reading uses Bessel's inequality $\sum_{j=1}^{\infty}$$|\langle x,x_{\alpha_{j}}\rangle|$$^{2}$$\leq$$\left\lVert x\right\rVert$ to conclude ($S_{n}$)$_{n}$ is Cauchy. How is this done?

Answer 1

if $P_n = \sum_{i=1}^n |p_j|^2$ is bounded then $(P_n)_{n\in \mathbb{N}}$ converges (since it is monotone increasing). Consequently, it is Cauchy. If you apply this to your $p_j =|\langle x, x_{a_j}\rangle|$, use the Bessel inequality and write down what this means you get your result.