Convergence of series $\sum_{n=1}^{\infty} \frac{\ln^\beta(n)}{n^\alpha}$

Question

I know if $\beta = 0$ it converges $\iff$ $\alpha > 1$. Also, it doesn't converge if $\alpha = 0$. I don't know what test to apply for the rest of cases. Any hints?

You can use the integral test to find the answer, which is explained here — Jean-Claude Arbaut, May 12 '14 at 16:23
For integer $\beta$, this is the $\beta^{th}$ derivative of the $\zeta$ function. — , May 12 '14 at 16:26
In fact, the name is Bertrand series. http://www.sosmath.com/calculus/series/bertrand/bertrand.html — , May 12 '14 at 16:46
I don't this should be marked as a duplicate, because this question includes the case (in the other question) where $\beta < 0$, whereas that question strictly assumes that $\beta \ge 0$. — Chill2Macht, Feb 12 '19 at 02:12

score 1 · Answer 1 · answered May 12 '14 at 16:55

For $\epsilon>0$ we have $1<\ln(n) <n^\epsilon $ for almost all $n$, therefore (replacing $\epsilon$ with $\frac{\epsilon}{|\beta|}$ if $\beta\ne 0$) $$n^{-\epsilon} < \ln^\beta(n)<n^\epsilon $$ for almost all $n$. Thus for $\beta>0$ and $\alpha>1$ we have convergence (pick $\epsilon<\alpha-1$) and for $\beta<0$ and $\alpha<1$ we have divergence.

Convergence of series $\sum_{n=1}^{\infty} \frac{\ln^\beta(n)}{n^\alpha}$

1 Answers1