$\sum\limits_{n=1}^\infty a_n^2$ and $\sum\limits_{n=1}^\infty b_n^2$ converge show $\sum\limits_{n=1}^\infty a_n b_n$ converges absolutely

Question

Given the series $\sum\limits_{n=1}^\infty a_n^2$ and $\sum\limits_{n=1}^\infty b_n^2$ converge. Show that the series $\sum\limits_{n=1}^\infty a_n b_n$ converges absolutely.

My idea so far:

• It's quite quite obvious that both given series converge absolutely
• So the Cauchy-Produc tells me that $\sum\limits_{n=1}^\infty a_n^2 b_n^2 = \sum\limits_{n=1}^\infty (a_n b_n)^2$ converges absolutely

I got stuck at that point. Can somehow give me a hint how to solve this ?

Thanks in advance!

Answer 1

\begin{align} 0 & \le (a+b)^2 = a^2 + b^2 + 2ab \\ 0 & \le (a-b)^2 = a^2 + b^2 - 2ab \\[10pt] \text{Therefore} \\ -2ab & \le a^2+b^2, \\ 2ab & \le a^2 + b^2, \\[10pt] \text{and consequently} \\ 2|ab| & \le a^2 + b^2. \end{align}

So $$ \sum_n |a_n b_n| \le \frac 1 2 \left( \sum_n a_n^2 + \sum_n b_n^2 \right) < \infty. $$

Answer 2

$\sum |a_n b_n|\le \sum b_n^2 + \sum a_n^2$

spoiler:

if $|a_n|\le|b_n|$ then $|a_n\cdot b_n|\le b_n^2$ else $|a_n\cdot b_n|\le a_n^2$, so $\sum |a_n b_n|\le \sum b_n^2 + \sum a_n^2$

Edit: Second spoiler (complete solution with all details):

if $|a_n|\le|b_n|$ then $|a_n\cdot b_n|\le b_n^2\le b_n^2+a_n^2$ else $|a_n\cdot b_n|\le a_n^2\le a_n^2+b_n^2$. In all cases $|a_n\cdot b_n|\le b_n^2+a_n^2$ so $\sum |a_n b_n|\le \sum (b_n^2 + a_n^2)=\sum b_n^2 + \sum a_n^2<\infty$.

Answer 3

HINT:

From Cauchy-Schwarz we have

$$\sum_{n=1}^{N}\left|a_n\, b_n\right|\le\sqrt{\left(\sum_{n=1}^{N}a_n^2\right)\,\left(\sum_{n=1}^{N}b_n^2\right)}$$