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

Not a duplicate, for $2$ reasons:

  • I cannot use the integral test
  • Other questions ask specifically when both $\alpha, \beta > 0$, on which we easily can apply Cauchy condensation criterion

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 $\alpha = 1, \beta > 1$

However, I'm struggling with the case that $\alpha > 0$ and $ \beta < 0$. Any hints?

  • $\begingroup$ 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 $\endgroup$ – User8128 Dec 12 '17 at 16:24
  • $\begingroup$ And @User8128 how would I show that the sequence of terms is decreasing? (otherwise I can't apply that criterion) $\endgroup$ – user370967 Dec 12 '17 at 16:43
  • $\begingroup$ 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). $\endgroup$ – User8128 Dec 12 '17 at 17:16
  • $\begingroup$ This doesn't seem obvious to me, since $\ln^{\beta}$ is decreasing for negative $\beta$ $\endgroup$ – user370967 Dec 12 '17 at 17:25
  • $\begingroup$ But $n^\alpha$ grows faster than $\ln^\beta(n)$ decreases (this is the classical result that $n^\epsilon \gg \ln^k(n)$ for any $\epsilon, k > 0$) $\endgroup$ – User8128 Dec 12 '17 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

  • $\begingroup$ And how would I show the case $\alpha < 0, \beta > 0$? $\endgroup$ – user370967 Dec 12 '17 at 16:45
  • $\begingroup$ @Math_QED see the edit relaod your page $\endgroup$ – Guy Fsone Dec 12 '17 at 16:48
  • $\begingroup$ 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? $\endgroup$ – user370967 Dec 12 '17 at 16:54
  • $\begingroup$ in that case it converges was that part of your question? $\endgroup$ – Guy Fsone Dec 12 '17 at 16:56
  • $\begingroup$ I accidentally pressed enter too early. $\endgroup$ – user370967 Dec 12 '17 at 16:56

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