TITLE: show that $\lim_{n \to \infty} n\int_0^1 e^{-tn}(\frac{\sinh t}{t})^n \, dt=1$
QUESTION [7 upvotes]: I tried to solve $$\lim_{n \to \infty} n\int_0^1\ldots \int_0^1\frac{1}{x_1+x_2+\ldots+x_n}\,dx_1\ldots dx_n$$
I started with 
$$ \frac{1}{x_1+x_2+\ldots+x_n}=\int_0^{\infty}e^{-t(x_1+x_2+\ldots+x_n)}\,dt$$
$$\Rightarrow I=\int_0^1\ldots\int_0^1\frac{1}{x_1+x_2+\ldots+x_n}dx_1\ldots dx_n=\int_0^{\infty}\int_0^1\ldots\int_0^1e^{-t(x_1+x_2+\ldots+x_n)}dx_1\ldots dx_n\,dt$$
$$\Rightarrow I=\int_0^{\infty}\int_0^1e^{-t(x_1)}dx_1\int_0^1e^{-t(x_2)} \, dx_2 \ldots\int_0^1e^{-t(x_n)}dx_n\,dt$$
$$\Rightarrow I=\int_0^\infty \left ( \frac{1-e^{-t}}{t}\right )^n\,dt$$
so our lim should be
$$\Rightarrow \lim_{n \to \infty} n\int_0^{\infty} \left (\frac{1-e^{-t}} t \right )^n \, dt$$
and with some steps :
$$\Rightarrow \lim_{n \to \infty} 2n\int_0^\infty e^{-tn}\left (\frac{\sinh t} t \right )^n\,dt$$
using mathematica i got numerically that the result is $2$.
i could see that $$\lim_{n \to \infty} \int_1^\infty e^{-tn}\left (\frac{\sinh t}{t}\right )^n\,dt=0$$
so how to prove that 
$$\lim_{n \to \infty} n\int_0 ^{1} e^{-tn}\left (\frac{\sinh t}{t}\right )^n\,dt=1$$

REPLY [2 votes]: For $x\in[-1,1]$,
$$
\begin{align}
\frac1{x^2}\left(\frac{\sinh(x)}{x}-1\right)
&=\sum_{k=1}^\infty\frac{x^{2k-2}}{(2k+1)!}\\
&\le\sum_{k=1}^\infty\frac1{6^k}\\[3pt]
&=\frac15\tag{1}
\end{align}
$$
Therefore, on $[0,1]$, $1\le\frac{\sinh(x)}{x}\le1+\frac{x^2}5\le e^{x^2/5}$.
$$
\begin{align}
&\lim_{n\to\infty}n\int_0^1e^{-tn}\left(\frac{\sinh(t)}{t}\right)^n\,\mathrm{d}t\\
&=\lim_{n\to\infty}\int_0^ne^{-t}\left(\frac{\sinh(t/n)}{t/n}\right)^n\,\mathrm{d}t\\
&=\color{#090}{\lim_{n\to\infty}\int_0^ke^{-t}\left[1,e^{\frac{t^2}{5n}}\right]\,\mathrm{d}t}
+\color{#C00}{\lim_{n\to\infty}\int_k^ne^{-t}\left[1,e^{\frac{t}{5}}\right]\,\mathrm{d}t}\\[6pt]
&=\color{#090}{1-e^{-k}}+\color{#C00}{O\!\left(e^{-4k/5}\right)}\\[12pt]
&=1+O\!\left(e^{-4k/5}\right)\tag{2}
\end{align}
$$
Since $(2)$ holds for all $k\ge0$, we have
$$
\lim_{n\to\infty}n\int_0^1e^{-tn}\left(\frac{\sinh(t)}{t}\right)^n\,\mathrm{d}t=1\tag{3}
$$