• Prove that the series $\sum_{n=1}^\infty {|a_n b_n|}$ and $\sum_{n=1}^\infty {(a_n + b_n)^2}$ converges
  • If the series $\sum_{n=1}^\infty {a^2_n}$ and $\sum_{n=1}^\infty {b^2_n}$ converges.

The part $\sum_{n=1}^\infty {(a_n + b_n)^2}$ does not necessarily converge, I had thought so, but I am doubting the first condition. Someone can help me, please.

  • 1
    $\begingroup$ Hint: Cauchy–Schwarz inequality. $\endgroup$
    – Gary
    Jul 11, 2020 at 19:57
  • $\begingroup$ I think $AM\ge GM $ on $a^2_n, b^2_n$ and then comparision test will work $\endgroup$ Jul 11, 2020 at 19:58

1 Answer 1


If the series $\sum_{n=1}^{\infty}a^{2}_{n}$ and $\sum_{n=1}b^{2}_{n}$ converge, then $\sum_{n=1}^{\infty}(a_{n}+b_{n})^{2}$ converges due to the comparison test. Indeed, one has that

\begin{align*} (a_{n} + b_{n})^{2} = a^{2}_{n} + 2a_{n}b_{n} + b^{2}_{n} \leq a^{2}_{n} + 2|a_{n}b_{n}| + b^{2}_{n} \leq 2(a^{2}_{n} + b^{2}_{n}) \end{align*}

where it has been used the AM-GM inequality: \begin{align*} \frac{a^{2}_{n} + b^{2}_{n}}{2}\geq \sqrt{a^{2}_{n}b^{2}_{n}} = |a_{n}b_{n}| \end{align*}

Similarly, the comparison test works for the other series, and we are done.

Hopefully this helps.

  • $\begingroup$ Less precise an upper bound, but still sufficient, is this: Since $\{\max( |a|,|b|),\min(|a|,|b|)\}=\{|a|,|b|\},$ we have $|ab|=|a| \cdot |b| \le$ $\le (\max(|a|,|b|))^2 \le$ $ (\max(|a|,|b|))^2+(\min(|a|,|b|))^2=$ $|a|^2+|b|^2.$ $\endgroup$ Jul 12, 2020 at 6:07

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