Find a constant $c_0 > 0$ that satisfies the following properties:
(i) The convergence of every series $\sum a_n$ with $a_n \geq 0$ implies the convergence of $\sum\limits_{n=1}^\infty b_n := \sum\limits_{n=1}^\infty \frac{\sqrt{a_n}}{n^c}$ for all $c > c_0$
(ii) There exists a convergent series $\sum a_n$ with $a_n \geq 0$ such that the series $\sum b_n$ diverges if $c = c_0$
My attempt:
$\sum\limits_{n=1}^\infty \frac{\sqrt{a_n}}{n^c} \leq \sum\limits_{n=1}^\infty \frac{\sqrt{M}}{n^c}$ since $a_n \geq 0$ is a convergent sequence and every convergent sequence is bounded. Note that this sum converges only when $c > 1$. So, we get that $c_0 = 1$. But from Rudin Chapter 3 problem 7, we know that the convergence of $\sum a_n$ automatically implies the convergence of $\sum \frac{\sqrt{a_n}}{n}$ for $a_n \geq 0$, so the bound of $c_0 = 1$ is not tight enough. Can anyone tell me what $c_0$ should be?
Any help would be appreciated!
Let $c_0=1/2$. To show $\sum_n b_n$ converges for $c=1/2+\epsilon,$ use AM-GM: $$\sqrt{\frac{a_n}{n^{1+2\epsilon}}} \le \frac{a_n}{2} +\frac{1}{2n^{1+2\epsilon}}.$$ For the case where $c=1/2,$ consider $a_n=\frac{1}{n\log^2(n+1)}.$