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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.


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

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.


(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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