The problem is this:
Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum \frac{\sqrt{a_n}}{n},$$ if $a_n\geq 0.$
I'm using the Cauchy's method:
$\sum \frac{\sqrt{a_n}}{n}$ converges $\iff$ $\sum2^k\frac{\sqrt{a_{2^k}}}{2^k}=\sum\sqrt{a_{2^k}}$ converges.
If $\sum a_n$ converges $\implies$ $\sum \sqrt{a_{2^k}}$ converges is true, I'm done.
Since $\sum a_n$ converges $\implies$ $a_n\to 0$, at least we have $\sqrt{a_{2^k}}\to 0$. But the thing is, near $0$, squarerooting makes it greater. So I can't use comparison test.
But we are choosing $a_{2^k}$ with great distance betwwen them, it will decay much faster. Like $2^k$ faster. So I'm highly confident that it will converge. However, I can't find $N$ such that $n>N$ implies $\sqrt{a_{2^n}}\leq a_n.$ Actually, I think there would be a counterexample (so the title is).
How do I proceed from here? And is there a counterexample?
Best Answer
The problem that you want to solve is easy to solve using the Cauchy-Schwarz inequality and the fact that the series $\sum_{n=1}^\infty\frac1{n^2}$ converge.
On the other hand, your approach is wrong, since the Cauchy condensation test assumes that he sequence $(a_n)_{n\in\mathbb N}$ is non-increasing, and you are not assuming that.
Finally, if$$a_n=\begin{cases}\frac1{k^2}&\text{ if $n=2^k$ for some $k$}\\0&\text{ otherwise,}\end{cases}$$then $\sum_{n=1}^\infty a_n$ converges, but $\sum_{k=0}^\infty\sqrt{a_{2^k}}$ diverges.