I'm reading Nadir Jeevanjee's An Introduction to Tensors and Group Theory for Physicists (2ed). In it, he says the following:
For $L^{2}([-a,a])$ for which $\{e^{i\frac{n \pi x}{a}}\}_{n\in \mathbb{Z}}$ is an orthonormal basis, ... any $f\in L^{2}([-a,a])$ can be written as \begin{equation} f=\sum\limits_{n=-\infty}^\infty c_{n}e^{i\frac{n \pi x}{a}} \qquad (2.37) \end{equation} where \begin{equation} \frac{1}{2a}\int\limits_{-a}^{a} |f|^2 dx = \sum\limits_{n=-\infty}^\infty |c_{n}|^2 < \infty \qquad (2.38) \end{equation} The converse to this is also true, and this is where the completeness of $L^{2}([-a,a])$ is essential: if a set of numbers $c_n$ satisfy $(2.38)$, then the series \begin{equation} g(x)=\sum\limits_{n=-\infty}^\infty c_{n}e^{i\frac{n \pi x}{a}} \qquad (2.39) \end{equation} converges, yielding a square-integrable function $g$.
I don't understand why if a set of numbers $c_n$ satisfy $(2.38)$, then g converges. Can someone please explain why?
(I suppose that if it converges, then by the completeness of $L^2$, we can say that it converges to an element of $L^2$.)
In general, if $v_n$ is some sequence of orthonormal vectors indexed by a set $U$ in a Hilbert space (a complete inner product space) and $\sum_{n\in U} |c_n|^2 < \infty$ then $$\sum_{n\in U} c_nv_n$$ converges.
Exactly how you prove this depends a bit on how you defined an infinite sum, but the basic idea stays the same: Suppose that you have some finite set of indices $I$ and some larger finite set $I'$. Define $$S_I=\sum_{n\in I}c_nv_n$$ $$S_{I'}=\sum_{n\in I'}c_nv_n.$$ Let's figure out how far apart $S_I$ and $S_{I'}$ are - note that $$S_{I'}-S_{I}=\sum_{n\in I'\setminus I}c_nv_n.$$ If you take the inner product of this (finite) sum with itself, you note that $\langle v_i,v_j\rangle = 0$ if $i\neq j$ and $1$ otherwise by the hypothesis of orthonormality. Thus $$\|S_{I'}-S_{I}\|^2 = \sum_{n\in I'\setminus I}|c_n|^2 \leq \sum_{n\in U\setminus I}|c_n|^2.$$ However, the fact that $\sum_{n\in U}|c_n|^2 < \infty$ implies that for any $\varepsilon>0$, there is a finite set $I$ such that $\sum_{n\in U\setminus I}|c_n|^2 < \varepsilon$ - meaning that the "tail" of the sum is small. In particular, if we set $\varepsilon = \alpha^2$, we would conclude that there is some finite set $I$ such that for every finite $I'$ containing $I$ we get $$\|S_{I'}-S_I\| < \alpha.$$ In the unlikely event that you are working with nets to define sums, this is exactly what Cauchy means; in the more likely even that you mean $$\sum_{n=-\infty}^{\infty}c_nv_n = \lim_{N\rightarrow\infty}\sum_{n=-N}^Nc_nv_n$$ the same fact implies that there is some $N$ such that if $N_1,N_2$ are greater than $N$ then $$\left\|\sum_{n=-N_1}^{N_1}c_nv_n - \sum_{n=-N_2}^{N_2}c_nv_n\right\| < 2\alpha$$ which implies that the sequence of values $\sum_{n=-N}^Nc_nv_n$ is a Cauchy sequence and hence converges.