TITLE: A Gamma limit $\lim_{n\rightarrow+\infty}\sum_{k=1}^n \left( \Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}$
QUESTION [25 upvotes]: Show that $$\lim_{n\rightarrow+\infty}\sum_{k=1}^n \displaystyle \left( \Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}$$
where $\gamma$ is the Euler-Mascheroni Constant.

Motivation : One can show that $$\lim_{n\rightarrow+\infty}\displaystyle\left(n-\Gamma\bigl(\frac{1}{n}\bigr)\right)=\gamma.$$ This means that $\Gamma\bigl(\frac{1}{n}\bigr)\sim n$ when $n$ is large. So we have that (even if is not correct) $\Gamma\bigl(\frac{k}{n}\bigr)\sim \frac{n}{k}$. It implies that $$\sum_{k=1}^{n}\displaystyle\left(\Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}\sim \sum_{k=1}^{n}(\frac{k}{n})^k.$$ Since the limit of the right sum exists and its value is $\frac{e}{e-1}$. Numerical calculations show that the limit of the sum involving the Gamma function would be $\frac{e^\gamma}{e^\gamma− 1}$.

REPLY [5 votes]: Since $\log(\Gamma(x))$ is convex, $\log(\Gamma(x))\ge-\gamma(x-1)$ and $\log(\Gamma(x))\ge(1-\gamma)(x-2)$. Thus, $\log(\Gamma(x))\ge-\gamma+\gamma^2$. That is, $\Gamma(x)\ge e^{-\gamma+\gamma^2}=0.78345806514\gt\frac34$.

For $1\le k\le\frac23n$,
$$
\begin{align}
\Gamma\left(\frac kn\right)^{-k}
&=\left(\frac kn\right)^k\,\Gamma\left(1+\frac kn\right)^{-k}\\
&\le\left(\frac{4k}{3n}\right)^k\tag1
\end{align}
$$
Since $\left(1+\frac1k\right)^k\lt e\lt\left(1+\frac1k\right)^{k+1}$, we have
$$
\frac4{3n}ke\le\frac4{3n}\frac{(k+1)^{k+1}}{k^k}\le\frac4{3n}(k+1)e\tag2
$$
Therefore, $\left(\frac{4k}{3n}\right)^k$ decreases while $\frac kn\lt\frac{3}{4e}$, then it increases.
Thus, applying $(1)$ and $(2)$,
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^{2n/3}\Gamma\left(\frac kn\right)^{-k}
&\le\lim_{n\to\infty}\overbrace{\ \ \ \ \frac4{3n}\ \ \ \ \vphantom{\left(\frac89\right)^{\!\frac23n}}}^{k=1}+\overbrace{\frac23n\max\!\left(\frac{64}{9n^2},\left(\frac89\right)^{\!\frac23n}\right)}^{2\le k\le\frac23n}\\
&=0\tag3
\end{align}
$$

For $0\le k\le\frac13n$, since $\Gamma\left(\frac{n-k}n\right)\ge1+\frac{\gamma k}n$ and $\left(1+\frac{\gamma k}n\right)^{n+\gamma k}\ge e^{\gamma k}$,
$$
\begin{align}
\Gamma\left(\frac{n-k}n\right)^{k-n}
&\le\left(1+\frac{\gamma k}n\right)^{k-n}\\[3pt]
&\le e^{-\gamma k\frac{n-k}{n+\gamma k}}\\[9pt]
&\le e^{-\frac{2\gamma}{3+\gamma}k}\tag4
\end{align}
$$
Furthermore, since $\Gamma\left(\frac{n-k}n\right)=1+\frac{\gamma k}n+O\left(\frac kn\right)^2$,
$$
\lim_{n\to\infty}\Gamma\left(\frac{n-k}n\right)^{k-n}=e^{-\gamma k}\tag5
$$
by Dominated Convergence, $(4)$ and $(5)$ show that
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=2n/3}^n\Gamma\left(\frac kn\right)^{-k}
&=\lim_{n\to\infty}\sum_{k=0}^{n/3}\Gamma\left(\frac{n-k}n\right)^{k-n}\\
&=\sum_{k=0}^\infty e^{-\gamma k}\\[3pt]
&=\frac{e^\gamma}{e^\gamma-1}\tag6
\end{align}
$$

Putting $(3)$ and $(6)$ together, we get
$$
\lim_{n\to\infty}\sum_{k=1}^n\Gamma\left(\frac kn\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}\tag7
$$