The series that I need to check is $2^{-\ln n}$ and $3^{-\ln n}$.
For the second case, after doing the preliminary test (which gave me zero so that I may proceed further), I tried to solve it by direct comparison test with another series $3^{-n}$, which is a converging series, and since the subsequent terms in the former are smaller than in the latter, I concluded that, the series $3^{-\ln n}$ converges as well.
But doing the same thing with the series $2^{-\ln n}$ and comparing it with $2^{-n}$, I again find it a converging series which turns out to be wrong.
Can you help me with it?
P.S-- Sorry for not using Latex as I don't know how to use it!
But what exactly is $\frac1{2^{\ln n}}$? $$2^{\ln n}=e^{\ln2\ln n}=n^{\ln2}$$ Thus $$\sum_{n=1}^\infty\frac1{2^{\ln n}}=\sum_{n=1}^\infty\frac1{n^{\ln2}}$$ and since $\ln 2<1$, the series diverges.
The series where $2$ is replaced with $3$ converges since $\ln3>1$, and the sum is denoted $\zeta(\ln3)$.