# Prove that the series $\sum_{n=1}^\infty {|a_n b_n|}$ and $\sum_{n=1}^\infty {(a_n + b_n)^2}$ converges

• 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.

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

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