# $\sum \sqrt{a_n b_n}$ converges when $\sum a_n$ and $\sum b_n$ converge

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.

• See this. Commented Feb 26, 2014 at 19:45
• 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. Commented Feb 26, 2014 at 19:46
• @DavidMitra So I can bound each term by a term of a converging series... so simple! Commented Feb 26, 2014 at 19:51
• @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? Commented Feb 26, 2014 at 19:53
• @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. Commented Feb 26, 2014 at 20:00

$$ab\le\frac12(a^2+b^2)$$
You could use a variant of the Cauchy-Schwartz inequality, which states that: $$\left(\sum_{n\in S} x_ny_n\right)^2\le\sum_{n\in S}(x_n)^2\sum_{n\in S}(y_n)^2$$Let $$x_n=\sqrt{a_n}$$ and $$y_n=\sqrt{b_n}$$. Then:$$\left(\sum_{n\in S} \sqrt{a_nb_n}\right)^2\le\sum_{n\in S}a_n\sum_{n\in S}b_n$$And since the sums $$\sum_{n\in S}a_n$$ and $$\sum_{n\in S}b_n$$ converge, your sum is bounded.