# Convergence of $\sum_{i=1}^{\infty} a_ib_i$ using Cauchy-Schwarz inequality

Let $$a_n,b_n$$ be complex numbers, and suppose that $$\sum_{i=1}^{\infty}|a_i|^2$$, $$\sum_{i=1}^{\infty}|b_i|^2$$ converge. Then, we have from Cauchy-Schwarz Inequality $$|\sum_{i=1}^{\infty} a_ib_i|^2\leq (\sum_{i=1}^{\infty}|a_i|^2)(\sum_{i=1}^{\infty}|b_i|^2),$$ so $$|\sum_{i=1}^{\infty} a_ib_i|$$ converges. But, does it imply that $$\sum_{i=1}^{\infty} a_ib_i$$ converges? If I can prove that $$\sum_{i=1}^{\infty} |a_ib_i|$$ converges, then we are done. But I don't know how.

• The point is that $\sum|a_ib_i|$ converges. Sep 18, 2020 at 11:39
• @AnginaSeng ... because? Sep 18, 2020 at 11:41
• @James2020 Let $A_i = |a_i|$ and $B_i = |b_i|$, then $$0 \le \left(\sum |a_i b_i|\right)^2 = \left(\sum A_i B_i\right)^2 \le \sum |A_i|^2 \sum |B_i|^2 = \sum A_i^2 \sum B_i^2 = \sum |a_i|^2 \sum |b_i|^2 < \infty.$$ Sep 18, 2020 at 11:43
• @CameronWilliams Where does the second inequality sign come from? Is this from Cauchy Schwartz inquality for real numbers? Sep 18, 2020 at 11:45
• The inequality $\left(\sum A_i B_i\right)^2 \le \sum |A_i|^2 \sum |B_i|^2$ ? That's simply the Cauchy-Schwarz inequality as you've depicted above. Sep 18, 2020 at 11:46

So that we can use Cauchy-Schwarz directly, let $$A_i = |a_i|$$ and $$B_i = |b_i|$$, then noting that $$A_i, B_i > 0$$,
$$0 \le \left(\sum_{i=1}^{\infty} |a_i b_i|\right)^2 = \left(\sum_{i=1}^{\infty} A_i B_i\right)^2 \le \sum_{i=1}^{\infty} A_i^2 \sum_{i=1}^{\infty} B_i^2 = \sum_{i=1}^{\infty} |a_i|^2 \sum_{i=1}^{\infty} |b_i|^2 < \infty.$$
The sequence $$S_n = \sum_{i=1}^n |a_i b_i|$$ is an increasing sequence of real numbers which is bounded above and thus converges. Hence $$\sum_{i=1}^{\infty} |a_i b_i|$$ exists and is finite and thus $$\sum_{i=1}^{\infty} a_i b_i$$ exists and is finite (absolute convergence implies conditional convergence).