# If $(V,\left<.,.\right>)$ is hilbert, show that $\sum_{k=1}^\infty \left<v,e_i\right>e_i=v$

Let $(V,\left<.,.\right>)$ a Hilbert space of infinite dimension and let $\{e_i\}_{i=1}^\infty$ an orthonormal basis. Show that $\left(\sum_{k=1}^n \left<v,e_i\right>e_i\right)_{n\in\mathbb N^*}$ is a Cauchy sequence and that it converge to $v$.

I tried as follow $$\left\|\sum_{k=n}^m \left<v,e_i\right>e_i\right\|^2=\left<\sum_{k=n}^m \left<v,e_i\right>e_i,\sum_{\ell=n}^m \left<v,e_i\right>e_i\right>=\sum_{k=n}^m \left<v,e_k\right>^2\underset{Cauchy-Schwarz}{\leq} \|v\|^2(m-n+1).$$

But I can't conclude.

For the fact that the limit is $v$, I juste set $$v=\sum_{i=1}^\infty v_ie_i$$ (am I sure that it converge or I have to show that the serie converge ?). Then, $$\left<v,e_i\right>=\left<\sum_{k=1}^n v_ke_k,e_i\right>\underset{(*)}{=}\sum_{k=1}^\infty v_k\left<e_k,e_i\right>=v_i.$$

My problem is to prove that the equality $(*)$ is correct. I know that for finite sum is correct, but how can I be sure for infinite sum ?

Let $s_n=\sum_{k=1}^n\left<v,e_k\right>e_k$. Then $$\left<v,s_n\right>=\sum_{k=1}^n|\left<v,e_k\right>|^2$$ and $$\|s_n\|^2=\sum_{k=1}^n|\left<v,e_k\right>|^2.$$ Thus $$\|s_n\|^2=\left<v,s_n\right>$$ and hence $$\|s_n\|^2\le \|v\|\|s_n\|$$ So $\|s_n\|\le\|v\|$ and hence $\sum_{k=1}^n|\left<v,e_k\right>|^2\le\|v\|^2$. Thus $\{s_n\}$ converges. It is easy to show $\lim_{n\to\infty}s_n=v.$