How do I go about proving that Square Root of the Product of Convergent Series Converges, where both are greater than 0. So $\sum \sqrt{x_n y_n}$, where $\sum x_n$ and $\sum y_n$ converge, and each $x_n$ and $y_n$ are greater than 0
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3$\begingroup$ Do you know the AM-GM inequality? $\endgroup$– JimmyK4542Commented Dec 18, 2014 at 8:53
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$\begingroup$ I do not. Never heard of it. $\endgroup$– user196910Commented Dec 18, 2014 at 8:57
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2$\begingroup$ $0\le(a-b)^2$.${}$ $\endgroup$– David MitraCommented Dec 18, 2014 at 9:01
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$\begingroup$ I am not sure what the hint $0≤(a−b)^2$ means $\endgroup$– user196910Commented Dec 18, 2014 at 10:34
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$\begingroup$ So, $0\le a^2+b^2-2ab\Rightarrow 2ab\le a^2+b^2$. Use $a=\sqrt {x_n}$, $b=\sqrt {y_n}$ and the Squeeze Theorem. $\endgroup$– David MitraCommented Dec 18, 2014 at 12:10
1 Answer
One way of going about this is to apply the Cauchy-Schwarz inequality. Looking at the partial sums $\displaystyle s_k = \sum_{j=1}^k \sqrt{x_jy_j}$, we have the following:
$$s_k^2 \leq \sum_{j=1}^k x_j \cdot \sum_{j=1}^k y_j$$
The partial sums $x_k$ and $y_k$ are bounded above by $\displaystyle L = \sum_{n=1}^\infty x_n$ and $\displaystyle M = \sum_{n=1}^\infty y_n$ respectively, since we know those series converge. Therefore, from the above, we get:
$$s_k^2 \leq L \cdot M$$
That is to say, $s_k \leq \sqrt{L \cdot M}$.
And so the sequence of partial sums $s_k$ is increasing and bounded above, so what can we conclude?
As Jimmy mentioned in the comments, the AM-GM inequality will get you there a lot faster. The inequality says that the geometric mean of a set of numbers is dominated by the arithmetic mean of that set. Applied to this specific case, we have, for all $k \in \mathbb{N}$:
$$\sqrt{x_ky_k} \leq \frac{x_k + y_k}{2}$$
And so we have a nice comparison test:
$$\sum \sqrt{x_ky_k} \leq \frac{1}{2}\sum \left( x_k + y_k \right) = \frac{1}{2} \left( \sum x_k + \sum y_k \right)$$