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I am interested in the convergence of the following sum: $\displaystyle \sum_{n = 1}^\infty \frac{1}{(\log n)^{n^p}}$ for $p \ge 0$. Clearly the sum passes the $n$th term test for convergence, is to complicated for ratio test. Applying the root test I get $\frac{1}{(\log n)^{n^{p-1}}}$ which I'm not sure is helpful. How should I proceed?

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  • $\begingroup$ $np$ OR $n^p$ ? $\endgroup$ – Empty Nov 6 '15 at 10:45
  • $\begingroup$ Note that $n^p > \log n$, and $(\log n)^{\log n} = n^{\log \log n}$. $\endgroup$ – Daniel Fischer Nov 6 '15 at 10:46
  • $\begingroup$ It is $n^p$, sorry for any confusion $\endgroup$ – MathMajor Nov 6 '15 at 10:46
  • $\begingroup$ @DanielFischer I get $\frac{1}{n^{\log \log n}}$. How would I finish this off? Comparison to $p$-series? $\endgroup$ – MathMajor Nov 6 '15 at 11:04
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    $\begingroup$ See here. $\endgroup$ – Daniel Fischer Nov 6 '15 at 11:06
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If $p=0 $ we have $$\sum_{n\geq2}\frac{1}{\log\left(n\right)}\geq\sum_{n\geq2}\frac{1}{n}=\infty $$ if $p>0 $ we note that $$\log^{n^{p}}\left(n\right)=e^{n^{p}\log\left(\log\left(n\right)\right)}\geq e^{n^{p}}\geq n^{a} $$ for any fixed $a>1$ for a sufficiently large $n$, $n\geq N $ say. So $$\sum_{n\geq N}\frac{1}{\log^{n^{p}}\left(n\right)}\leq\sum_{n\geq N}\frac{1}{n^{a}}<\infty. $$

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Hint :

Use Cauchy's condensation test.

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