Question: Let $\{a_i\}_{i=1}^\infty$ and $\{b_i\}_{i=1}^\infty$ be in $\mathbb{C}^\infty$. Show $\sum_{i=1}^\infty a_i^{\ast}b_i$ is convergent, using in particular the Schwarz inequality.
Attempt at answer:
$\{a_i\}_{i=1}^\infty$, $\{b_i\}_{i=1}^\infty$ both in $C^\infty$, which is the set of all complex sequences. A property of $C^\infty$ says that a complex sequence $|a\rangle =\{\alpha_i\}_{i=1}^\infty$ such that $\sum\limits_{i=1}^\infty |\alpha_i|^2$ is finite.
So, let $\{\alpha_i\}_{i=1}^\infty$ and $\{\beta_i\}_{i=1}^\infty$ be sequences in $C^\infty$. Then $\sum\limits_{i=1}^\infty |\alpha_i|^2$ and $\sum\limits_{i=1}^\infty |\beta_i|^2$ are both finite, meaning $|a>$ and $|b>$ are finite, which implies that $|\langle a|b\rangle|^2$ is finite. Then, by the Schwarz Inequality, we have $|\langle a|b\rangle|^2\leq$ $\langle a|a\rangle \langle b|b\rangle$ also finite.
Then, since $|a\rangle $ is finite, then $|a^\ast \rangle$ is finite. So, $ \langle a^\ast|a^\ast \rangle \langle b|b\rangle$ is finite.
Thus, $\sum\limits_{i=1}^\infty \alpha_i^\ast\beta_i$ is finite, and so convergent.
Hint: The Cauchy-Schwarz Inequality says $$ \sum_k|a_kb_k|\le\left(\sum_k|a_k|^2\right)^{1/2}\left(\sum_k|b_k|^2\right)^{1/2} $$