Convergence of $\sum_{n=1}^\infty \frac{\ln^k(n) }{n^a}$

Question

Show when $\displaystyle \sum_{n=1}^\infty \frac{\ln^k(n) }{n^a}$ is convergent.

Tried using convergence tests. I tried to calculate $$\lim: \lim_{n\to \infty} \frac{\frac{\ln^k(n) }{n^a}}{\frac{1}{n^a}}$$ but that doesn't help me. Need some advises, thanks

Answer 1

HINT

We should declare the range for $a$ and $k$, otherwise we need to consider the following cases:

• For $a>1$ take $b=\frac{1+a}2$ and apply limit comparison test with $\sum \frac1{n^b}$.
• For $0\le a<1$ and $k\ge0$ use comparison test with $\sum \frac1{n^a}$.
• For $0\le a<1$ and $k<0$ refer to Cauchy condensation test.
• For $a<0$ we have that $a_n \not \to 0$.

Note also that when $k<0$ we need to start from $n=2$ in order to have a well defined expression.