Does the series $$\sum_{n=1}^{\infty} \frac{a^{n+k} n !}{n^{n+k}}$$ with $a>0$ and $k \in \mathbb{N}$


Using the ratio test already found that it converges for ae, but couldn't find another method for a=e, as the limit in the ratio test goes to 1.


closed as off-topic by Martin R, José Carlos Santos, blub, mrtaurho, Shailesh Apr 17 at 2:48

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By Stirling approximation:

$$\frac{a^{n+k}n!}{n^{n+k}}\sim \left(\frac{a}{n}\right)^{n+k}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n=u_n$$

$$u_n = \left(\frac{a}{e}\right)^n \sqrt{2\pi}a^k n^{1/2-k}$$

As you remarked, if $a>e$ this series diverges, and converges if $a<e$. If $a=e$, you therefore have

$$u_n = C n^{1/2-k}$$ with $C$ a constant, which converges only if $k-1/2>1$, that is, $k>3/2$.


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