# Prove convergence of $\sum_{n=1}^{\infty} a^{\ln n}$ for $0<a<\frac{1}{e}$ [duplicate]

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

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 $$0, I get $$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 CusterOct 10 '18 at 16:04

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• Hint. $a^{\log n} = n^{\log a}$. – Sangchul Lee Oct 10 '18 at 14:35

## 2 Answers

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. 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.