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.
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.
