A Gamma limit $\lim_{n\rightarrow+\infty}\sum_{k=1}^n \left( \Gamma\bigl(\frac{k}{n}\bigr)\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}$ 
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 $\frac{e^\gamma}{e^\gamma− 1}$.
 A: 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
$$
\lim_{n\to\infty}\sum_{k=1}^n\Gamma\left(\frac kn\right)^{-k}=\frac{e^\gamma}{e^\gamma-1}\tag7
$$
A: 
Step 1. If 
  $$I_1(n)=\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$
  Then $\lim\limits_{n\to\infty}I_1(n)=0$.

Proof. Indeed, since $\Gamma$ is decreasing on $(0,1]$ we have
$$
I_1(n)\leq\sum_{1\leq k\leq\sqrt{n}}\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}\leq\sum_{k=1}^\infty\left(\Gamma\left(\frac{1}{\sqrt{n}}\right)\right)^{-k}=\frac{1}{\Gamma(1/\sqrt{n})-1}$$
and step 1. follows.

Step 2. If 
  $$I_2(n)=\sum_{\sqrt{n}<k\leq n/2}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$
  Then $\lim\limits_{n\to\infty}I_2(n)=0$.

Proof. Recall that $\Gamma$ attains its minimum $\approx0.8856$, on $[1,2]$, at some some point $x_0\approx1.4616$. In particular, $\Gamma(x)\geq2/3$ for $1\leq x\leq 2$. So, for $\sqrt{n}<k\leq n/2$ we have
$$
\frac{k}{n}\Gamma\left(\frac{k}{n}\right)=\Gamma\left(1+\frac{k}{n}\right)
\geq\frac{2}{3}
$$
Thus,  for $\sqrt{n}<k\leq n/2$, we have $\Gamma(k/n)>4/3$. It follows that
$$
I_2(n)\leq \sum_{k>\sqrt{n}}\left(\frac{3}{4}\right)^k=4\left(\frac{3}{4}\right)^{\lceil\sqrt{n}\rceil}
$$
and step 2. follows.

Step 3. If 
  $$I_3(n)=\sum_{n/2<k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}$$
  Then $\lim\limits_{n\to\infty}I_3(n)=\dfrac{e^\gamma}{e^\gamma-1}$.
  where $\gamma$ is the Euler-Mascheroni constant.

Proof.
Note first that, with $p=n-k$,
$$
I_3(n)=\sum_{0\leq p<n/2}\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n}
=\sum_{p=0}^\infty a_p(n)
$$
with
$$a_p(n)=\left\{\matrix{\left(\Gamma\left(1-\frac{p}{n}\right)\right)^{p-n}&\hbox{if}& 0\leq p<n/2\cr0&\hbox{otherwise}}\right.$$
Now, since $\Gamma(1)=1$ and $\Gamma'(1)=-\gamma$ we have, for a fixed $p$ and large $n$:
$$(p-n)\ln\Gamma\left(1-\frac{p}{n}\right)=(p-n)\ln\left(1+\frac{\gamma p}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)=-\gamma p+\mathcal{O}\left(\frac{1}{n}\right)$$
Thus
$$
\forall\,p\geq 0,\quad \lim_{n\to\infty}a_p(n)=e^{-\gamma p}.\tag{1}
$$
Now, we will need the next lemma.
Lemma. For $t\in[1/2,1]$ we have $(\Gamma(t))^{t/(1-t)}\geq \Gamma(1/2)=\sqrt{\pi}.$
Taking, this lemma for granted, we conclude by taking $t=1-p/n$ when $0\leq p<n/2$, that 
$$
\forall\,p\geq 0,n\geq 1,\quad  a_p(n)\leq \left(\frac{1}{\sqrt{\pi}}\right)^p.
\tag{2}
$$
and clearly, $$\sum_{p=0}^\infty \left(\frac{1}{\sqrt{\pi}}\right)^p<+\infty\tag{3}$$  Combining $(1)$, $(2)$ and $(3)$ we conclude that
$$
\lim_{n\to\infty}I_3(n)=\lim_{n\to\infty}\sum_{p=0}^\infty a_p(n)
=\sum_{p=0}^\infty\lim_{n\to\infty}a_p(n)=
\sum_{p=0}^\infty e^{-\gamma p}=\frac{e^\gamma}{e^\gamma-1}.$$
The desired conclusion follows:
$$
\lim_{n\to\infty}\sum_{1\leq k\leq n}\left(\Gamma\left(\frac{k}{n}\right)\right)^{-k}=
\lim_{n\to\infty}(I_1(n)+I_2(n)+I_3(n))=\frac{e^\gamma}{e^\gamma-1}.
$$
Proof of the Lemma. Let $f(t)=\dfrac{t}{1-t}\ln\Gamma(t)$.
Then $f'(t)=\dfrac{g(t)}{(1-t)^2}$ with
$$g(t)=\ln\Gamma(t)+t(1-t)\psi(t);\quad\hbox{where $\psi(t)=\Gamma'(t)/\Gamma(t)$}$$
and $g'(t)=(1-t)h(t)$ with
$$h(t)=2\psi(t)+t\psi'(t)$$
and finally $h'(t)=3\psi'(t)+t\psi''(t)=\sum_{k=0}^\infty\frac{3k+t}{(k+t)^3}>0$.
So, $h$ is increasing, and $\lim_{t\to0^+}h(t)=-\infty$, $h(1)=\frac{\pi^2}{6}-2\gamma>0$. This proves that $h(t)<0$ for $0<t<x_0$ and $h(t)>0$ for $x_0<t<1$, for some $x_0$.
And $g$ is decreasing on $[0,x_0]$ and increasing on $[x_0,1]$. But 
$\lim_{t\to0^+}g(t)=+\infty$, $g(1)=0$. This proves that $g$ has exactly one change of sign on $(0,1)$ from positive to negative. This proves that the minimum of $f$ on $[1/2,1]$ is $\min(f(1/2),f(1))=f(1/2)$, and the lemma is proved.
