Given any orthonormal sequence $(e_k)$ in a Hilbert space $H$, we may consider series of the form $$\Large \sum_{k=1}^{\infty} c_ke_k.$$
I must prove that $$\Large \sum_{k=1}^{\infty} c_ke_k\in H\implies c_k=(x,e_k),$$ where $x$ denotes the sum of the series. I found the proof of this statement on Kreyszig, but I did not understand the final part. I report it in full.
Taking the inner product of $s_n$ and $e_j$ ad using the orthonotmality, we have $$(s_n, e_j)=c_j\quad\text{for}\quad j=1,\dots, k,$$ where $k\le n$ fixed. By assumption $s_n\to x$. Since the inner product is continuous we have $$c_j=(s_n, e_j)\to (x, e_j)\quad (j\le k).$$
Now here's the part that I didn't understand, or rather, it doesn't seem written in detail. I don't like proof where things are said in words:
Here we can take $k\le n$ as large as we please because $n\to \infty$, so we have $c_j=(x, e_j)$ for every $j=1, 2, \dots$
Some simple steps are omitted? Could someone explain to me better? Thank you.
$\sum_{k=1}^{\infty}c_k e_k \in \mathcal{H}$ is not a very clear notation. What the author intends is that $\lim_{N\rightarrow\infty}\sum_{k=1}^{N}c_ke_k$ exists. A second problem is that $c_k$ is being used as a scalar. You want to distinguish vectors and scalars, and the standard convention is that $a,b,c,d,\cdots$ are vectors and that $\alpha,\beta,\gamma,\delta,\cdots$ are scalars.
What does it mean to say that $\lim_{N\rightarrow\infty}\sum_{n=1}^{N} \alpha_n e_n$ exists and equals $x$? It means that, for every given real $\epsilon > 0$, there exists a positive integer $M$ (that generally depends on $\epsilon$) such that $$ \|\sum_{n=1}^{N}\alpha_n e_n-x\| < \epsilon,\;\;\; \mbox{whenever} \; N > M. $$ In that case, it is standard to write $\sum_{n=1}^{\infty}\alpha_n e_n = x$, or $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\alpha_n e_n=x$. The inner product $\langle \cdot,\cdot\rangle$ is continuous in both coordinates because of the Cauchy-Schwarz inequality $$ |\langle x,y\rangle| \le \|x\|\|y\|. $$ Therefore, the following limits exist and are equal for all $y\in\mathcal{H}$: $$ \langle \sum_{n=1}^{\infty}\alpha_n e_n, y\rangle = \langle \lim_N\sum_{n=1}^{N}\alpha_n e_n,y\rangle =\lim_{N}\langle\sum_{n=1}^{N}\alpha_n e_n,y\rangle \\ =\lim_{N}\sum_{n=1}^{N}\alpha_n\langle e_n,y\rangle =\sum_{n=1}^{\infty}\alpha_n \langle e_n,y\rangle. $$ Therefore, $\langle \sum_{n=1}^{\infty}\alpha_n e_n,y\rangle=\sum_{n=1}^{\infty}\alpha_n\langle e_n,y\rangle$, provided the vector limit sum exists. In particular, if you set $y=e_k$, then $$ \langle \sum_{n=1}^{\infty}\alpha_n e_n,e_k \rangle=\sum_{n=1}^{\infty}\alpha_n\langle e_n,e_k\rangle = \alpha_k. $$