Find the nature of $\sum_{n = 1}^\infty \frac{1}{e \cdot \sqrt{e} \cdot \sqrt[3]{e} \cdots \sqrt[n]{e}}$

Question

I need to find whether the following series converges or diverges:

$$\sum_{n = 1}^\infty \frac{1}{e \cdot \sqrt{e} \cdot \sqrt[3]{e} \cdots \sqrt[n]{e}}$$

It seems to diverge, so I tried to use Bertrand's Test, but this involves computing the following limit:

$$\lim_{n \rightarrow \infty} e^{\frac{1}{n + 1}} n \ln n - (n + 1) \ln(n + 1)$$

Which, by graphing it, seems to be $-1$, but I don't know how to prove this. Can you help me find the nature of the series (maybe using another method)? Thanks!

Answer 1

Try rendering the denominator as $e^{1+(1/2)+(1/3)+...(1/n)}$. You should know that the sum in the exponent tends to $\log n+\text{ a constant}$ as $n\to\infty$, from which you can identify a well-known comparison series.