How to test the convergence of $\sum^{\infty}_{n=2}\frac{1}{\ln{n}}$ and $\sum^{\infty}_{n=2}\frac{\ln{n}}{n^{1.1}}$? How to test the convergence of $\sum^{\infty}_{n=2}\frac{1}{\ln{n}}$ and $\sum^{\infty}_{n=2}\frac{\ln{n}}{n^{1.1}}$? 
For the first one, I use basic comparison and compare it to $\frac{1}{n}$, since $\frac{1}{n}\lt \frac{1}{\ln{n}}$ and $\frac{1}{n}$ diverges, so $\sum^{\infty}_{n=2}\frac{1}{\ln{n}}$ diverges.
Foe the second one, I have no idea how to start with?
Any suggestion?

Use the Cauchy condensation test and note that $$2^na_{2^n}=\frac{2^n\log(2^n)}{(2^n)^{1.1}}=\frac{n\log2}{2^{0.1n}}$$ so that $\sum 2^n a_{2^n}$ converges by the ratio test and hence the original series converges.
