# Convergence of $\sum_{n = 1}^\infty \frac{1}{(\log n)^{n^p}}$ for $p \ge 0$.

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?

• $np$ OR $n^p$ ? – Empty Nov 6 '15 at 10:45
• Note that $n^p > \log n$, and $(\log n)^{\log n} = n^{\log \log n}$. – Daniel Fischer Nov 6 '15 at 10:46
• It is $n^p$, sorry for any confusion – MathMajor Nov 6 '15 at 10:46
• @DanielFischer I get $\frac{1}{n^{\log \log n}}$. How would I finish this off? Comparison to $p$-series? – MathMajor Nov 6 '15 at 11:04
• See here. – Daniel Fischer Nov 6 '15 at 11:06

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 $$\sum_{n\geq N}\frac{1}{\log^{n^{p}}\left(n\right)}\leq\sum_{n\geq N}\frac{1}{n^{a}}<\infty.$$