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

You should clarify the range for $a$ and $k$, otherwise we need to consider some cases. — user, Sep 01 '18 at 07:48
Possible duplicate of Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ — Nosrati, Sep 01 '18 at 10:06
@MaorRocky Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE — user, Oct 23 '18 at 21:15

user · Answer 1 · 2018-09-01T08:30:08.037

2

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.

edited Sep 01 '18 at 08:30

answered Sep 01 '18 at 07:23

user

154,566

Concrete and satisfying+ – Mikasa Sep 01 '18 at 08:15
1

@ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye! – user Sep 01 '18 at 08:19

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

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