Prove $\sum_{n=1}^{\infty}|a_{n}b_{n}|$ converges if $\sum_{n=1}^{\infty}a_{n}^{2}$ and $\sum_{n=1}^{\infty}b_{n}^{2}$ converge

Question

This is a homework problem for an undergrad topology course.

Let $l^{2}$ be the set of all real-valued sequences $(c_{n})$ where $\sum_{n=1}^{\infty}c_{n}^{2}$ converges. Let $(a_{n}),(b_{n})\in l^{2}$.

Claim: $\sum_{n=1}^{\infty}|a_{n}b_{n}|$ converges.

I've done a couple of examples where $a_{n}$ and $b_{n}$ are general harmonic series. I've also written out the definition of convergence for $a_{n}$ and $b_{n} $. My problem is that thus far I do not understand on any level why my claim is true.

Answer 1

No need to use Cauchy Bunyakovsky Schwarz here.

We have $(a_n+b_n)^2 = a_n^2+ b_n^2 + 2 a_nb_n\ge0 $, hence $a_n^2+ b_n^2 \ge - 2 a_nb_n$.

We have $(a_n-b_n)^2 = a_n^2+ b_n^2 - 2 a_nb_n\ge0 $, hence $a_n^2+ b_n^2 \ge 2 a_nb_n$.

Hence $a_n^2+ b_n^2 \ge 2 |a_nb_n|$.

Then $2 \sum_{k=0}^N |a_nb_n| \le \sum_{k=1}^\infty ( a_n^2+b_n^2) $ for all $N$ and the result follows.

Answer 2

By the Cauchy Schwarz inequality, $$\left(\sum_{n}a_n^2\right)^{1/2}\left(\sum_{n}b_n^2\right)^{1/2}\ge \left(\sum_{n}|a_nb_n|\right)$$ So we are done.

Answer 3

As mentioned before, by the Cauchy-Schwarz inequality $$\sum_{n=1}^\infty |a_nb_n| \leq \sqrt{\big(\sum_{n=1}^\infty a_n^2\big)} \sqrt{\big(\sum_{n=1}^\infty b_n^2\big)}$$

and you know the series inside the roots converge to real numbers by assumption. However, you may have to prove this statement. A hint for that would be to square both sides of this inequality, and then show the result by induction.