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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$?

There are other questions on MSE which ask about this series, but this question is distinct because

  • 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

Xander Henderson
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    The Cauchy Condensation test works (in fact, this is sort of the canonical example of when to use the Cauchy Condensation test). https://en.wikipedia.org/wiki/Cauchy_condensation_test – User8128 Dec 12 '17 at 16:24
  • And @User8128 how would I show that the sequence of terms is decreasing? (otherwise I can't apply that criterion) –  Dec 12 '17 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). – User8128 Dec 12 '17 at 17:16
  • This doesn't seem obvious to me, since $\ln^{\beta}$ is decreasing for negative $\beta$ –  Dec 12 '17 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 $\epsilon, k > 0$) – User8128 Dec 12 '17 at 17:32
  • I suppose in the case that $\alpha > 1$ or $\alpha < 1$, you don't even need the condensation test, you can show convergence\divergence by comparison with $1/n^{\alpha \pm \delta}$ for sufficiently small $\delta$. When $\alpha = 1$ and $\beta \le 0$, you can show divergence by comparison with $1/n$. Then we are in the case when $\alpha =1$ and $\beta > 0$ whence it is obvious that the terms decrease and you can apply the condensation test. – User8128 Dec 12 '17 at 17:39
  • Related : https://math.stackexchange.com/questions/1264767/infinite-sum-of-dfrac1np-lnnq – Arnaud D. Dec 03 '19 at 11:07

1 Answers1

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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

Guy Fsone
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