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
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$\begingroup$ Note that $n^p > \log n$, and $(\log n)^{\log n} = n^{\log \log n}$. $\endgroup$ – Daniel Fischer♦ Nov 6 '15 at 10:46
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$\begingroup$ It is $n^p$, sorry for any confusion $\endgroup$ – MathMajor Nov 6 '15 at 10:46
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$\begingroup$ @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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1$\begingroup$ 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 $$\sum_{n\geq N}\frac{1}{\log^{n^{p}}\left(n\right)}\leq\sum_{n\geq N}\frac{1}{n^{a}}<\infty. $$
answered Nov 6 '15 at 11:11
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$\begingroup$ $\sum 1/n^p$ is divergent for $0<p\le 1$. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 9 '15 at 7:15
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Hint :
Use Cauchy's condensation test.
