TITLE: Limit $\lim_{x\to\infty}\left(\sum_{n=1}^{\infty}\left(\frac{x}{n}\right)^n \right)^{1/x}$
QUESTION [10 upvotes]: Evaluate $$\lim_{x\to\infty}\left(\sum_{n=1}^{\infty}\left(\frac{x}{n}\right)^n \right)^{1/x}$$
My attempt: I tried to expand the internal series by substituting the consecutive values of n, but the series that appears is not known to me. Kindly help me to solve the problem.

REPLY [7 votes]: In this answer, it is shown, using Bernoulli's Inequality, that
$$
\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}\tag1
$$
Thus,
$$
\begin{align}
\frac{x}{(n+1)e}\le\frac{x}{n+1}\left(\frac{n}{n+1}\right)^n=\frac{\left(\frac x{n+1}\right)^{n+1}}{\left(\frac xn\right)^n}
=\frac{x}{n}\left(\frac{n}{n+1}\right)^{n+1}\le\frac{x}{ne}\tag2
\end{align}
$$
Therefore, by induction, for $n\ge1$,
$$
\frac{e}{n!}\left(\frac xe\right)^n\le\left(\frac xn\right)^n\le\frac{x}{(n-1)!}\left(\frac xe\right)^{n-1}\tag3
$$
Summing $(3)$, we get
$$
e\left(e^{x/e}-1\right)\le\sum_{n=1}^\infty\left(\frac xn\right)^n\le xe^{x/e}\tag4
$$
and raising $(4)$ to the $1/x$ power yields
$$
\left(e\left(1-e^{-x/e}\right)\right)^{1/x}e^{1/e}\le\left(\sum_{n=1}^\infty\left(\frac xn\right)^n\right)^{1/x}\le x^{1/x}e^{1/e}\tag5
$$
and by the Squeeze Theorem, we have
$$
\lim_{x\to\infty}\left(\sum_{n=1}^\infty\left(\frac xn\right)^n\right)^{1/x}=e^{1/e}\tag6
$$