Infinite sum convergence proof Question: Is my proof of the theorem below vaild?
Theorem: 
If $\sum a^2_k $ and $\sum b^2_k $ both converge, then $\sum a_k b_k$ converges.
My proof: 
Assume $\sum a^2_k $ and $\sum b^2_k $ both converge.
There are two cases, namely $\sum a^2_k \le \sum b^2_k$ or $\sum a^2_k \ge \sum b^2_k$.
$\sum a^2_k \le \sum b^2_k \implies \vec a \cdot \vec a \le \vec b \cdot \vec b \implies \lvert \vec a \rvert^2 \le \lvert \vec b \rvert^2 \implies \lvert \vec a \rvert \le \lvert \vec b \rvert$ 
$\sum a_k b_k = \vec a \cdot \vec b \le \lvert \vec a \rvert \lvert \vec b \rvert \le \lvert \vec b \rvert \lvert \vec b \rvert = \vec b \cdot \vec b = \sum b_k^2$
By the comparison test, $\sum a_k b_k$ converges.
Similarly, $\sum a_k b_k$ converges when assuming $\sum a^2_k \ge \sum b^2_k$.
Hence, if $\sum a^2_k $ and $\sum b^2_k $ both converge, then $\sum a_k b_k$ converges. Q.E.D
 A: One way is to observe that $2|a_kb_k|\le a_k^2+b_k^2$ (AM-GM) now $\sum a_kb_k$ converges by comparison test.

I would avoid using a vector $\vec a = (a_1,a_2,\cdots)$ (which does not have a finite dimension). Especially when we have an easier solution.
If you want to improve your solution, you can invoke Cauchy's criterion to get a finite sum $\sum_{i=n}^{m}{a_i^2}$ and similarly for $b_k$ and apply the same procedure (here the vector is of finite dimension).
A: Proof:
Assume that $\sum a_k^2$ and $\sum b_k^2$ both converge.
$0\le(\lvert a_k\rvert - \lvert b_k \rvert)^2 = \lvert a_k\rvert^2 - 2\lvert a_k\rvert \lvert b_k\rvert + \lvert b_k\rvert^2 = a_k^2 - 2\lvert a_k b_k\rvert + b_k^2$
$\implies 2\lvert a_k b_k \rvert \le a_k^2 + b_k^2\implies \lvert a_k b_k \rvert \le a_k^2 + b_k^2$
$\{\lvert a_k b_k \rvert\}_{k=0}^\infty$ and $\{a_k^2 + b_k^2\}_{k=0}^\infty$ are two positive sequences.
$\sum (a_k^2+b_k^2) = \sum a_k^2 + \sum b_k^2$ converges.
Therefore, by the comparison test, $\sum \lvert a_k b_k \rvert$ converges.
By definition, $\sum a_k b_k $ is absolutely convergent.
Hence, $\sum a_k b_k $ converges. Q.E.D
