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I'm trying to proof the convergence of the series

$\sum\limits_{n=1}^{\infty} \exp\left(- \dfrac{n^k}{\log(n)} \right)$

where $0 < k < \frac{1}{2}$ is a positive constant.

I tried to use the ratio test, but it was inconclusive. I think I can show the convergence using the comparision test, but I can't seem to find a convergent majorant.

Does anyone know a convergent majorant or another method to show the convergence? ;) Thanks in advance!

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    $\begingroup$ Index should $n$ should start at $2$ since $log(1) = 0$. $\endgroup$ – MathNewbie Jun 10 '15 at 20:37
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As a first step, note that $n^k/\log n = \Omega(n^{k/2})$, and so the summand is $\exp - \Omega(n^{k/2})$ (here $k/2$ is an arbitrary number in $(0,k)$). Second, $e^{-x} = O(1/x^t)$ for any $t>0$. In particular, choosing $t = 4/k$ (or any other $t > 2/k$), we get that the summand is $O(1/n^2)$. Since $\sum_{n=1}^\infty 1/n^2$ converges, so does your series.

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Note that $$\exp\left(\frac{n^{k}}{\log\left(n\right)}\right)>n^{\alpha}\Leftrightarrow n^{k}>\alpha\log^{2}\left(n\right) $$ and this is true for all $k>0 $, for all $\alpha $ and sufficiently large $n $. Then, if we take $\alpha=1+\epsilon,\,\epsilon>0 $ we have that exists some $N\in\mathbb{N} $ such that $$\sum_{n\geq N}\exp\left(-\frac{n^{k}}{\log\left(n\right)}\right)\leq\sum_{n\geq N}\frac{1}{n^{\alpha}}<\infty. $$

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