For which $p's \in \mathbb{R}_\geq1$ does the series $\sum _{n\in \mathbb{N}}\Bigl(\frac{1}{\sqrt{n}\log(1+n)}\Bigr)^{p}$ converge

Thoughts Clearly the sequence $\frac{1}{\sqrt{n}\log(1+n)}$ tends to $0$ as $n$ tends to infinity, although that doesn't mean that much as we know $\frac{1}{n}$ diverges. I've tried a few convergence tests such as the ratio test and found r=1 so that doesn't help, I've tried the integral test but can't see a way to integrate this function so I'm not really sure what to do now.

  • $\begingroup$ Have you tried Cauchy condensation test? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Oct 7 '18 at 15:17
  • $\begingroup$ Comparison test would do all the work, maybe. The standards are $\sum 1/n^k$ for $k>1$ and $\sum 1/n$. $\endgroup$ – xbh Oct 7 '18 at 15:23
  • $\begingroup$ Okay so the comparison test shows that its either converges for all p's or it's diveges for all p's, going to try Cauchy condensation test for p=1 now. $\endgroup$ – Roger Oct 7 '18 at 15:40
  • $\begingroup$ No, the series converges for some $p,$ diverges for some $p.$ $\endgroup$ – zhw. Oct 7 '18 at 15:51
  • $\begingroup$ Can you explain why this is wrong then $\frac{\Bigg(\frac{1}{\sqrt{n}\log(1+n)}\Bigg)^{p}}{\Bigg(\frac{1}{\sqrt{n}\log(1+n)}\Bigg)^{1}}=\Bigg(\frac{1}{\sqrt{n}\log(1+n)}\Bigg)^{p-1}$ which goes to $1 $ if $ p=1$ and $o$ if $\infty>p>1$ so by the comparison test I only need to look at $p=1$ $\endgroup$ – Roger Oct 7 '18 at 16:05


Near $\infty$, the general term is equivalent to $\dfrac1{n^{p/2}\log^pn}$, which is a Bertrand's series.

Now, for the general Bertrand's series $\;\displaystyle\sum_{n=2}^\infty\frac1{n^\alpha (\log n)^\beta}$, it is known that it converges if and only if

  • $\alpha>1$, or
  • $\alpha=1$ and $\beta>1$.
  • $\begingroup$ Okay this makes it pretty clear that it converges for $p \geq 2$ still not 100% sure what I did wrong with the comparison test. $\endgroup$ – Roger Oct 7 '18 at 16:19
  • $\begingroup$ The problem is probably that the test to apply depends on the value of $p$. It's also the case for Bertrand's series: ffor $\alpha>1$ the comparison test works fine, and for `alpha=1$, we need the integral test. $\endgroup$ – Bernard Oct 7 '18 at 16:26

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