Series with log and exponential $\sum_{n=1}^{\infty} \frac{\log n}{n^a}$

Question

How could this series be solved: $$\sum_{n=1}^{\infty} \frac{\log n}{n^a}$$

I was thinking of comparing it with the convergent series $\frac1{n^b}$ where b>1 and using the comparison test but I get stuck. Is there any other way to do it?

Answer 1

Note that the series

$$\sum \frac{\log n}{n^a}$$

• for $a\le 0$ diverges since $a_n\to +\infty$
• for $0< a\le 1$ diverges by comparison with $$\sum \frac1{n^a}$$
• for $a> 1$ converges by comparison with $$\sum \frac1{n^{\frac{a+1}2}}$$