How to test the convergence of $\sum^{\infty}_{n=2}\frac{1}{\ln{n}}$ and $\sum^{\infty}_{n=2}\frac{\ln{n}}{n^{1.1}}$?

Question

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?

Answer 1

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.

Answer 2

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.