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 $n^{\alpha}-$test with $\alpha = 1.05$. 
The $n^{\alpha}-$test says that if $\alpha > 1$, and $a_n \ge0$ eventually, and if:
$$\lim n^{\alpha}a_n = 0$$
Then $\sum a_n$ converges.
