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

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.

• Have you tried Cauchy condensation test? – GNUSupporter 8964民主女神 地下教會 Oct 7 '18 at 15:17
• Comparison test would do all the work, maybe. The standards are $\sum 1/n^k$ for $k>1$ and $\sum 1/n$. – xbh Oct 7 '18 at 15:23
• 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. – Roger Oct 7 '18 at 15:40
• No, the series converges for some $p,$ diverges for some $p.$ – zhw. Oct 7 '18 at 15:51
• 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 $p=1$ – 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 $$\beta>1$$.
• 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. – Roger Oct 7 '18 at 16:19
• 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. – Bernard Oct 7 '18 at 16:26