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

Let $$(x_{\alpha})_{\alpha \in I}$$ be an orthonormal family of vectors in a Hilbert space H and let $$J:=$${$$\alpha$$ $$\inI$$ $$|$$ $$\langlex$$,$$x_{\alpha}\rangle\neq0$$} 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?

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.