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

Assume for the moment the truth of the following statement $$x\int_{0}^{1} \dfrac{1}{t^{tx}}dt = \sum_{n \geq 1} \left(\dfrac{x}{n}\right)^n. \label{e:1} \tag{*}$$

Since $\lim_{x \to \infty} x^{1/x} = 1$ the answer is $$\displaystyle \lim_{x \to \infty} \left(\int_{0}^{1} \left(f(t)\right)^x dt\right)^{1/x} $$ where $f(t) = \dfrac{1}{t^{t}}$, assuming the limit exists.

Since $f(t)$ is continuous on $[0,1]$ (defining $0^0 = 1$) and using the well known result, that for a positive continuous function $g$ on $[0,1]$ we have $$\lim_{x \to \infty}\left(\int_{0}^{1} \left(g(t)\right)^x\right)^{1/x} = \sup_{t \in [0,1]} g(t) \label{e:2}\tag{+}$$ it follows that required limit is $ \sup_{t \in [0,1]} f(t)$. It is easy to see that the maximum of $f(t)$ occurs at $1/e$ so the answer is $f(1/e)=e^{1/e}$.

For a proof of $\eqref{e:1}$ we start with the identity $$ \int_{0}^{\infty} \exp( -\lambda u) u^{\alpha - 1}du = \dfrac{\Gamma(\alpha)}{\lambda^\alpha}. $$ Put $\lambda = \dfrac{n}{x}, \alpha = n$ to get
$$ \int_{0}^{\infty} \dfrac{1}{\Gamma(n)}\exp(-n\dfrac{u}{x}) u^{n-1}du = \left(\dfrac{x}{n}\right)^n $$

So $$ \begin{align} \sum_{n \ge 1}\left( \dfrac{x}{n} \right)^n &= \int_{0}^{\infty}\exp(-\dfrac{u}{x}) \sum_{n \geq 1} \dfrac{\left(u \exp(-\dfrac{u}{x})\right)^{n-1}}{\Gamma(n)} du\\ &= \int_{0}^{\infty} \exp\left(-\dfrac{u}{x} + u\exp(-\dfrac{u}{x})\right)du. \end{align} $$

Substituting $u = -x \log t$ in the above integral we have $$\sum_{n \ge 1}\left( \dfrac{x}{n} \right)^n = \int_{0}^{1} \exp(\log t -x t\log t) x \dfrac{dt}{t} = x\int_{0}^{1} \dfrac{1}{t^{tx}}dt. $$

A proof for $\eqref{e:2}$.

Let $M = \sup_{x \in [0,1]} g(x) > 0.$ Clearly $\left(\int_{0}^{1} (g(t))^{x}\right)^{1/x} \leq (M^x)^{1/x} = M.$ There is a $x_0 \in [0,1]$ such that $M = g(x_0)$, and for any $ \epsilon > 0$ we have a $\delta > 0 $ such that $|x - x_0| \leq \delta$ and $0 \leq x \leq 1$ implies $ g(x) \geq M - \epsilon$ and $$ \left(\int_{0}^{1} (g(t))^{x}\right)^{1/x} \geq \left(\int_{|x - x_0| < \delta} (g(t))^{x}\right)^{1/x} \geq (M-\epsilon) \ell^{1/x} $$ where $\ell$ is the length of the interval where $0 \leq x \leq 1$ and $ |x - x_0| \leq \delta.$ Since $\ell^{1/x} \to 1$ as $x \to \infty$ the result follows.