I'm preparing for the preliminary exam, and some practice problems are all of the same form. One problem is to prove that convergence of $\sum a_n$ where $a_n \geq 0$ implies convergence of $\sum \frac{\sqrt{a_n}}{n}$. I know how to prove that case using Cauchy-Schwarz (or AM-GM) as is discussed here. Now I have another problem:
- Suppose $a_n \geq 0$ and the series $\sum a_n$ converges. Prove the series $\sum \frac{\sqrt{a_n}}{n^{2/3}}$ also converges.
- Give an example of a divergent series $\sum a_n$ such that the series $\sum \frac{\sqrt{a_n}}{n^{2/3}}$ converges.
By Abel's test it's easy to see that, in general, convergence at $r$ implies convergence at any $r' > r$.
I'm not sure how to go about it for $r = 2/3$. Maybe it follows from the case $r = 1$ but I don't see how.
Best Answer
Applying the Cauchy-Schwarz inequality
$$\left|\sum_{n=1}^N x_ny_n\right|\le \sqrt{\sum_{n=1}^N x_n^2\sum_{n=1}^N y_n^2}$$
to the case for which $x_n=\sqrt{a_n}$ and $y_n=\frac1{n^r}$, we find that
$$\sum_{n=1}^N \frac{\sqrt a_n}{n^r}\le \sqrt{\sum_{n=1}^N a_n \sum_{n=1}^N \frac1{n^{2r}}}\tag1$$
As $N\to \infty$, the first sum on the right-hand side of $(1)$ converges by assumption while the second converges whenever $r>1/2$. Inasmuch as $2/3>1/2$, the the series $\sum_{n=1}^\infty \frac{\sqrt a_n}{n^{2/3}}$ converges.
The condition that $\sum_{n=1}^\infty a_n$ converges is not a necessary one for the series $\sum_{n=1}^\infty \frac{\sqrt a_n}{n^r}$ to converge. For example, if $a_n=\frac1{n^p}$, $p<1$, then the series $\sum_{n=1}^\infty \frac{\sqrt a_n}{n^r}$ converges for $r>1-p/2$.