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

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 ?

• Why do you call this Cauchy product ? The way you wrote it is much simpler. Commented Dec 2, 2015 at 22:06
• From $(a-b)^2\ge 0$, one deduces $ab\le{1\over2}(a^2+b^2)$. You could use this and the Comparison Test. Commented Dec 2, 2015 at 22:11
• Mirko is is absolutely right, Cauchy product was reserved for something else, don't use it arbitrarily. Commented Dec 2, 2015 at 22:59

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

• Isn't the AM-GM inequality great? Commented Dec 2, 2015 at 23:58

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

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 $\sum |a_n b_n|\le \sum (b_n^2 + a_n^2)=\sum b_n^2 + \sum a_n^2<\infty$.

• I will up-vote this if you add what you earlier said in a comment. Commented Dec 2, 2015 at 22:26
• @MichaelHardy thanks Commented Dec 2, 2015 at 22:27
• @Dr.MV I added a second spoiler with all details to avoid any misinterpretation Commented Dec 3, 2015 at 2:33
• @Mirko Yes! That is better!!! Well done! Commented Dec 3, 2015 at 2:37
• @Mirko No worry. I hadn't read your added spoiler. So, apology. I hope your class went well and you enjoy teaching. - Mark Commented Dec 3, 2015 at 2:44

HINT:

From Cauchy-Schwarz we have

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