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I am to show that $\sum \sqrt{a_n b_n}$ converges when $\sum a_n$ and $\sum b_n$ converge (here the series are assumed to have non-negative terms). I am unsure how to approach this problem; since I don't know what the series would converge to, I tried using Cauchy's criterion. Hence my goal was to bound

$$ \sum_{i=n+1}^{n+k} \sqrt{a_i b_i}$$

for some $n$ large enough and any $k \geq 1.$ I tried to express this in terms of the $a_i$ and $b_i$ separately (to use convergence of the series $\sum a_n$ and $\sum b_n$) by writing the above expression as

$$\frac{1}{2} \left( \sum_{i=n+1}^{n+k} (\sqrt{a_i}+\sqrt{b_i})^2 - \sum_{i=n+1}^{n+k} a_i - \sum_{i=n+1}^{n+k} b_i \right),$$

but I'm not sure if this really helps.

Thanks for any help.

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  • $\begingroup$ See this. $\endgroup$ – David Mitra Feb 26 '14 at 19:45
  • $\begingroup$ If the terms are non-negative, the sequence of partial sums is monotonically nondecreasing. Thus it is convergent if and only if it is bounded. $\endgroup$ – Daniel Fischer Feb 26 '14 at 19:46
  • $\begingroup$ @DavidMitra So I can bound each term by a term of a converging series... so simple! $\endgroup$ – user131708 Feb 26 '14 at 19:51
  • $\begingroup$ @DanielFischer Does this help at all, if we don't think about using the AM-GM inequality? Or did you have something else in mind? $\endgroup$ – user131708 Feb 26 '14 at 19:53
  • $\begingroup$ @user131708 The next step would be AM-GM. The point is that you don't need to bother with the Cauchy criterion and $\sum\limits_{i=n+1}^{n+k}$ once you know that $\sqrt{a_nb_n} \leqslant \frac12(a_n+b_n)$ and the sum of the larger terms is finite. $\endgroup$ – Daniel Fischer Feb 26 '14 at 20:00
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Hint

$$ab\le\frac12(a^2+b^2)$$

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  • $\begingroup$ Nice, concise hint: AM-GM inequality. +1 $\endgroup$ – DonAntonio Feb 26 '14 at 19:52
  • $\begingroup$ Perfect (and so simple...)! Thanks, everyone. $\endgroup$ – user131708 Feb 26 '14 at 19:54

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