How does one show that $S_1=S_2$ and has the closed form of $1?$

Question

Consider these sum

$$\lim_{n\to \infty}\sum_{k=1}^{n}\gamma^k\left(n-\Gamma\left({k\over n}\right)\right)^{-k}=S_1$$ and $$\lim_{n\to \infty}\sum_{k=1}^{n}(-\gamma)^{1+k}\left(n-\Gamma\left({k\over n}\right)\right)^{-k}=S_2$$ Where $\gamma$ is Euler-Mascheroni Constant

How does one show that $S_1=S_2$ and has the closed form of $\color{red}1?$

Answer 1

Since the $\Gamma$ function has a simple pole at $z=0$, the behavior of $n-\Gamma\left(\frac{k}{n}\right)$ for $k\in[1,n]$ is pretty simple to understand. We have $$ n-\Gamma\left(\frac{1}{n}\right)=\gamma-\frac{1}{2n}\left(\zeta(2)+\gamma^2\right)+O\left(\frac{1}{n^2}\right) $$ while for $k\geq 2$ we have $$ \left(n-\Gamma\left(\frac{k}{n}\right)\right)^{-k} =O\left(\frac{1}{n^2}\right)$$ uniformly with respect to $k$. It follows that

$$ S_1 = \color{red}{1},\qquad S_2 = \color{red}{\gamma} $$

since the only contribute that really matters is the contribute given by $k=1$.