Behavior of $\sum_i^n (1-\frac{1}{i})^s$ as a function of $s$? I'm interested in behavior of the following sum as a function of $s$
$$\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s$$
For $n=1000$ and $s\in (1,10000)$, this seems almost linear on a log-plot (notebook), any tips how to model this analytically?


I shall scale $s$ with $n$ and write $s=\alpha n$ with some $\alpha>0$. It will be assumed that $\alpha>0$ is bounded and $n$ is large. First, from the monotonicity properties of the integrand,
\begin{align*}
\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx}  \le \sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^{\alpha n} } & \le \int_1^{n + 1} {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} \\ & \le \int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx}  + \left( {1 - \frac{1}{{n + 1}}} \right)^{\alpha n} ,
\end{align*}
i.e.,
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^{\alpha n} }  = \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx} .
$$
Now we write
\begin{align*}
&\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx}  = \int_{1/n}^1 {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}} 
\\ &
 = \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}}  + \int_{1/n^{2/3} }^1 {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}} .
\end{align*}
Here
$$
\int_{1/n^{2/3} }^1 {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}}  \le n^{2/3} \exp \left( {\alpha n\log \left( {1 - \frac{1}{{n^{2/3} }}} \right)} \right) \le n^{2/3} e^{ - \alpha n^{1/3} } ,
$$
and
\begin{align*}
& \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha n\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{{t^2 }}}  = \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha nt - \alpha n\mathcal{O}(t^2 )} \right)\frac{{dt}}{{t^2 }}} 
\\ & = \int_{1/n}^{1/n^{2/3} } {\exp \left( { - \alpha nt} \right)\frac{{dt}}{{t^2 }}} \left( {1 +\mathcal{O}\!\left( {\frac{1}{{n^{1/3} }}} \right)} \right) = n\alpha \int_\alpha ^{\alpha n^{1/3} } {e^{ - s} \frac{{ds}}{{s^2 }}} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{n^{1/3} }}} \right)} \right)
\end{align*}
for any fixed $\alpha>0$. Consequently,
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\int_1^n {\left( {1 - \frac{1}{x}} \right)^{\alpha n} dx}  = \alpha \int_\alpha ^{ + \infty } {e^{ - s} \frac{{ds}}{{s^2 }}}  = \alpha \Gamma ( - 1,\alpha )
$$
for any fixed $\alpha>0$. Here $\Gamma(a,z)$ is the upper incomplete gamma function. Finally,
$$\boxed{
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^{\alpha n} }  = \alpha \Gamma ( - 1,\alpha )
}$$
for any fixed $\alpha>0$. Note that in terms of the exponential integral, $\alpha \Gamma ( - 1,\alpha ) = e^{ - \alpha }  - \alpha E_1 (\alpha )$.
Try to plot, for example,
$$
\frac{1}{{1000}}\sum\limits_{i = 1}^{1000} {\left( {1 - \frac{1}{{i}}} \right)^s } \;\; \text{ and } \;\; \frac{s}{{1000}}\Gamma\! \left( { - 1,\frac{s}{{1000}}} \right)
$$
against each other on the range $1\leq s\leq 10000$.
