# Given $a_n > 0$ for all $n$ and $\sum a_n$ converges. Show that if $b > \frac{1}{2}$, then $\sum_{n=1}^\infty n^{-b} \sqrt{a_n}$ converges.

I attempted the integral test, limit comparison test, ratio test, and root test.

Limit comparison test: $\lim \sup \frac{n^{-b} \sqrt{a_n}}{a_n} < \infty$?

I get $\frac{0}{0}$ and apply l'Hopital's rule. (Note: I believe that when applying l'Hopital's rule, I take the derivative with respect to $n$, in which case, I suppose I can think of $a_n$ as $f(n)$, and the $\frac{\partial}{\partial x} f(n) = f'(n) \to 0$ since $a_n \to 0$.)

In most cases, I'm left with a perpetual loop of $\frac{0}{0}$.

I'm wondering if I should instead approach this problem via a comparison test and find some $\sum b_n$ that converges such that $0 \leq \sum_{n=1}^\infty n^{-b} \sqrt{a_n} \leq \sum b_n$.

Can this be proven using one of the aforementioned tests?

$$n^{-b}\sqrt{a_n} \leq \frac12(n^{-2b}+a_n)$$
• Nice. Let's say you are using the baby version of AM-GM $\sqrt{xy} \le (x+y)/2$.
• What if $b\leq 1/2$? The conclusion seems to be the following: there are examples of $(a_n)$ such that the series is either convergence or divergent. Mar 31 at 4:04