Problem
Prove Bessel's inequality using Lenma below.
Bessel's Inequality
Let $H$ be Hilbert space on $\mathbb C$, $(\cdot , \cdot)$ be an inner product in $H$, and $\{ e_k \}_{k\in \mathbb Z}$ be an orthonormal system in $H$.
Then, for all $x\in H,$ $$\sum_{k\in \mathbb Z}|(x, e_k)|^2\leqq \|x\|^2$$ holds.
Lenma
Let $(X, (\cdot , \cdot))$ be an inner product space on $\mathbb C$, and $\{ e_k \}_{k\in \mathbb Z}$ be an orthonormal system.
Then, for all $x\in X,$
$$\| x-\sum_{k=-N}^N (x, e_k)e_k\|^2 =\| x\|^2 -\sum_{k=-N}^N |(x, e_k)|^2 \ \mathrm{for \ all}\ N \in \mathbb N$$ holds.
(I have already proven this Lenma.)
I have to prove Bessel's inequality using this Lenma, but my approach has one problem.
My approach
Let $x\in H.$
Then, from Lenma, I get $$\| x-\sum_{k=-N}^N (x, e_k)e_k\|^2 =\| x\|^2 -\sum_{k=-N}^N |(x, e_k)|^2 \ \mathrm{for \ all}\ N \in \mathbb N$$
Since the LHS is nonnegative, I get $$\| x\|^2 \geqq \sum_{k=-N}^N |(x, e_k)|^2 \ \mathrm{for \ all}\ N \in \mathbb N.$$
Since this holds for all $N\in \mathbb N$, letting $N\to \infty$, $$\| x\|^2 \geqq \sum_{k\in \mathbb Z} |(x, e_k)|^2 .$$
But this has one problem. I didn't show $\displaystyle\lim_{N\to \infty}\sum_{k=-N}^N |(x, e_k)|^2 =\sum_{k\in \mathbb Z} |(x, e_k)|^2 $ exists.
To show this, I separate $\sum_{k=-N}^N |(x, e_k)|^2$ as $$\sum_{k=1}^N |(x, e_{-k})|^2+\sum_{k=1}^N|(x, e_k)|^2+|(x, e_0)|^2=\sum_{k=1}^N\left[ |(x, e_{-k})|^2+|(x, e_k)|^2\right]+|(x, e_0)|^2 $$ and I want to show $\sum_{k=1}^\infty\left[ |(x, e_{-k})|^2+|(x, e_k)|^2\right]$ exists.
Let $a_N:=\displaystyle\sum_{k=1}^N \left[ |(x, e_{-k})|^2+|(x, e_k)|^2\right]$, and it suffices to show that $\{a_n \}_{N=1}^\infty$ is Cauchy for the completeness of $H$.
For $N>M,$
$|a_N - a_M|= \left| \displaystyle\sum_{k=M+1}^N \left[ |(x, e_{-k})|^2+|(x, e_k)|^2\right]\right| $.
I cannot see this goes to $0$ as $M\to \infty.$ How can I see this ?
I would like you to give me any help.