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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!

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

3 Answers 3

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

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    $\begingroup$ Isn't the AM-GM inequality great? $\endgroup$
    – Mark Viola
    Dec 2, 2015 at 23:58
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$\sum |a_n b_n|\le \sum b_n^2 + \sum a_n^2$

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

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

From Cauchy-Schwarz we have

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

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