Show that $\sum_{n=1}^{\infty} a^{\ln~n}$ is convergent if and only if $0<a<\frac{1}{e}$.
Proof:
$\ln~n<n$ for all $n \geq 1$.Hence the necessary condition for the series to be convergent will be $\lim_{n \rightarrow \infty} a^{\ln~n}=0$. And this will happen if $0<a<1$. Please give me some hint about how to proceed further.
For $0<a<1$, $a^{\ln(n)}$ is a decreasing sequence. Hence by Cauchy condensation test, $$\sum a^{\ln(n)}\text{ is convergent} \iff \sum 2^na^{\ln(2^n)}\text{ is convergent} $$ The last sum is a geometric series, convergent for $0<a<1/e$.
Another approach would be to see that $a^{\ln(n)}=n^{\ln(a)}$, $\sum n^{\ln(a)}$ is a Dirichlet series, convergent only for $\ln(a)<-1$.