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.

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

1 Answer 1


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).


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