How do we know that the $\ell^2$ inner product converges? For vectors $a, b \in \ell^2$, the argument that I have seen for the convergence of the inner product is that the Cauchy-Schwarz inequality says that
$$
|(a, b)| = \left| \sum_{i = 1}^n \overline{a_i} b_i \right| \le \sqrt{ \sum_{i = 1}^n |a_i|^2 } \sqrt{ \sum_{i = 1}^n |b_i|^2 } = |a| |b|
$$
and the right side exists in the limit as $n \to \infty$, so the left side must exist in that limit as well.
But I am not convinced by that argument, because actually we have only shown that the left side is always less than or equal to the right side, not that it is equal. For example, $\sin(x) \le 1 + e^{-x}$ for all $x$, and the limit of $1 + e^{-x}$ as $x \to \infty$ exists, but the limit of $\sin(x)$ does not exist.
So how do we know that $\sum_{i = 1}^\infty \overline{a_i} b_i$ exists?
 A: Let $a,b\in \ell^2(\Bbb{N};\Bbb{C})$. For each $n\in \Bbb{N}$, we have
\begin{align}
\sum_{i=1}^n|\overline{a_i}b_i|&=\sum_{i=1}^n|a_i||b_i|\leq \sqrt{\sum_{i=1}^n|a_i|^2}\sqrt{\sum_{i=1}^n|b_i|^2}\leq \|a\|_2\|b\|_2<\infty.
\end{align}
So, $\{\sum_{i=1}^n|\overline{a_i}b_i|\}_{n=1}^{\infty}$ is a weakly-increasing, bounded sequence of non-negative numbers, so its limit as $n\to\infty$ exists (infact equals the supremum over all $n$). i.e $\sum_{n=1}^{\infty}|\overline{a_n}b_n|$ exists and is finite. Finally, absolute convergence of a series of complex numbers implies convergence of that series of complex numbers (this fact is actually equivalent to $\Bbb{C}$ being complete; and this holds more generally for any real/complex normed space as well: a normed vector space is complete if and only if every absolutely convergent series converges in the norm topology as well). Hence, $\sum_{n=1}^{\infty}\overline{a_n}b_n$ exists (i.e $\lim\limits_{n\to\infty}\sum_{i=1}^n\overline{a_i}b_i$ exists in $\Bbb{C}$).
