Limit $\lim_{x\to\infty}\left(\sum_{n=1}^{\infty}\left(\frac{x}{n}\right)^n \right)^{1/x}$

Stirling's formula gives you

$$ n^{-n} \sim \frac{\sqrt{2\pi n}}{n!e^n}. $$ Hence there exists positive constants $c_1$, $c_2$ such that

$$ \frac{c_1}{n!e^n} \leq n^{-n} \leq \frac{c_2}{(n-1)! e^n}. $$

Plugging these bounds into our sum yields respectively

$$ \begin{align} \sum_{n=1}^{\infty} \left(\frac{x}{n}\right)^n &\geq c_1\sum_{n=1}^{\infty} \frac{x^n}{n!e^n} \\ &=c_1(e^{x/e}-1) \end{align} $$ and $$ \begin{align} \sum_{n=1}^{\infty} \left(\frac{x}{n}\right)^n &\leq c_2\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!e^n} \\ &=c_2\frac{x}{e}e^{x/e}. \end{align} $$

Hence

$$\bigl(c_1(e^{x/e}-1)\bigr)^{1/x} \leq \left(\sum_{n=1}^{\infty} \left(\frac{x}{n}\right)^n\right)^{1/x} \leq \left(c_2\frac{x}{e}e^{x/e}\right)^{1/x}. $$

It is now clear that the limit seeked is $e^{1/e}$.

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 $$ \lim_{x\to\infty}\left(\sum_{n=1}^\infty\left(\frac xn\right)^n\right)^{1/x}=e^{1/e}\tag6 $$