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Prove the convergence of the series: $$\sum_{n=1}^{\infty} a^{\ln n},\,\text{for} \,\,0<a<\frac{1}{e}.$$

Attempt. I have proved the non-convergence in the case $a\geq 1/e$ (using the comparison test and getting $a^{\ln n}\geq \frac{1}{n}$). In case $0<a<\frac{1}{e}$, I get $a^{\ln n}<\frac{1}{n}$ and the above test doesn't work. Ratio test, root test are also not applicable here.

Thanks in advance for the help.

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marked as duplicate by Martin R, Matthew Towers, Brahadeesh, Christopher, Chris Custer Oct 10 '18 at 16:04

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    $\begingroup$ Hint. $a^{\log n} = n^{\log a}$. $\endgroup$ – Sangchul Lee Oct 10 '18 at 14:35
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Taking $a=e^{\ln a}$, we have

$$a^{\ln n}=\left(e^{\ln a}\right)^{\ln n}=e^{\ln a\ln n}=e^{\ln n\ln a}=n^{\ln a}$$

Since $\ln a$ is a constant when $a>0$, the p-series test tells us that the series converges iff $\ln a<-1$. This gives $a<e^{-1}=\frac{1}{e}$. This is the desired result.

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(Following the hint by @SangchulLee)

$$\sum_{n=1}^{\infty} \alpha^{\ln n}=\sum_{n=1}^{\infty}n^{\ln \alpha}= \sum_{n=1}^{\infty}\frac{1}{n^{-\ln\alpha}}$$ so we get a harmonic series with $-\ln\alpha>1$ for $0<\alpha<\frac{1}{e}$ and therefore convergence.

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