# On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$

Study the convergence of $$\sum_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$$ where $$\alpha, \beta \in \mathbb{R}$$

I have proved that:

1. This series diverges when $$\alpha \leq 0$$.

2. This series converges when $$\alpha > 1, \beta > 0$$

3. This series diverges when $$0 < \alpha < 1, \beta > 0$$

4. This series converges when $$\alpha = 1, \beta > 1$$

Question: What happens when $$\alpha > 0$$ and $$\beta < 0$$?

• I would like an argument which does not rely on the integral test for series convergence, and

• this question considers all real $$\alpha$$ and $$\beta$$, while other questions ask only about $$\alpha, \beta > 0$$, where we can apply Cauchy condensation criterion

• The Cauchy Condensation test works (in fact, this is sort of the canonical example of when to use the Cauchy Condensation test). en.wikipedia.org/wiki/Cauchy_condensation_test Commented Dec 12, 2017 at 16:24
• And @User8128 how would I show that the sequence of terms is decreasing? (otherwise I can't apply that criterion)
– user370967
Commented Dec 12, 2017 at 16:43
• It's clear the the terms are decreasing if you believe that $n^\alpha \ln^\beta(n)$ is increasing (which seems obvious enough that it can be stated without justification). Commented Dec 12, 2017 at 17:16
• This doesn't seem obvious to me, since $\ln^{\beta}$ is decreasing for negative $\beta$
– user370967
Commented Dec 12, 2017 at 17:25
• But $n^\alpha$ grows faster than $\ln^\beta(n)$ decreases (this is the classical result that $n^\epsilon \gg \ln^k(n)$ for any $\epsilon, k > 0$) Commented Dec 12, 2017 at 17:32

In that last case it diverges those are Bertrand's series see here: indeed we have

$$\frac{1}{n^{\alpha/2}\ln^\beta n} =\left[\frac{1}{\frac{2\beta}{\alpha}n^{\alpha/2\beta}\ln n^{\alpha/2\beta}}\right]^{\beta}\to \left[\frac{1}{\frac{2\beta}{\alpha}0^-}\right]^{\beta} =\infty$$ Since if $\alpha<0$ and $\beta>0$ then, $$\lim_{n\to\infty}n^{\alpha/2\beta}=0\implies \lim_{n\to\infty}n^{\alpha/2\beta}\ln n^{\alpha/2\beta} =0^-$$

Then there exists $N$ such that $n>N$ we have

$$\frac{1}{n^{\alpha/2}\ln^\beta n}>1\implies \frac{1}{n^{\alpha}\ln^\beta n}>n^{-\alpha/2}$$ That is $$\sum_{n=N}^{\infty}\frac{1}{n^{\alpha}\ln^\beta n}>\sum_{n=N}^{\infty}n^{-\alpha/2} =\infty$$ from this you get the divergence

• And how would I show the case $\alpha < 0, \beta > 0$?
– user370967
Commented Dec 12, 2017 at 16:45
• @Math_QED see the edit relaod your page Commented Dec 12, 2017 at 16:48
• Moreover, if $\alpha =1, \beta > 0$, we have: the series $\sum \frac{1}{n\ln^{\beta}(n)}$ and by cauchy's condensation criterion, we get $\sum \frac{1}{\ln^{\beta}(2) n}$ which diverges, but in the comments and sources on the internet it should converge for $\beta > 1$. Where am I going wrong?
– user370967
Commented Dec 12, 2017 at 16:54
• in that case it converges was that part of your question? Commented Dec 12, 2017 at 16:56
• I accidentally pressed enter too early.
– user370967
Commented Dec 12, 2017 at 16:56