Square Root of the Product of Convergent Series Converges

Question

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

Answer 1

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?

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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:

$$\sum \sqrt{x_ky_k} \leq \frac{1}{2}\sum \left( x_k + y_k \right) = \frac{1}{2} \left( \sum x_k + \sum y_k \right)$$