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?

 A: I assume you are interested in asymptotics.  It seems you are interested in both large $s$ and large $n$. The answer will be different depending on whether $s \ll n$ or $n \ll s$.
For $s \ll n$:
Since you are taking an average of an arbitrarily large sum, asymptotically any finite number of terms will not matter. This is enough to assume that $1/i \ll 1$. This allows you to use the approximation
$$(1+x)^s \approx 1+sx$$ which is valid for small $x$.  This is the tangent line approximation.
This gives
\begin{align*}
\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s &\approx \frac{1}{n}\sum_{i=1}^n \left(1-\frac{s}{i}\right) \\
&= 1 -  \frac{1}{n}\sum_{i=1}^n \frac{s}{i}
\end{align*}
Now use the further asymptotic approximation for the harmonic series:
$$\sum_{i=1}^n \frac{1}{i} \approx \log n$$
to get
\begin{align*}
1 -  \frac{1}{n}\sum_{i=1}^n \left(\frac{s}{i}\right) &\approx 1 -  \frac{s\log n}{n}.
\end{align*}
For $n \ll s$:
In this case, assuming we are considering $s$ to be arbitrarily large while $n$ is fixed, we eventually have
$$(1-1/2)^s \ll (1-1/3)^s \ll (1-1/4)^s \ll \cdots \ll (1-1/n)^s.$$
In this case the first $n-1$ terms become negligible and only the last term is dominant.  Then we get
$$\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s \approx \frac{1}{n} \left(1-\frac{1}{n}\right)^s$$
as $s\rightarrow \infty$.
For $0 \ll s \approx n$:
Of this I am not sure.  You may need to specify more carefully the relation between $s$ and $n$.
A: 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 $1\leq s\leq 10000$.
A: Since $\forall 1\leq i\leq n,\forall s\geq0, (1-{1\over i})^s\leq (1-{1\over n})^s$, so  $\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\leq\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{n}\right)^s=(1-{1\over n})^s$, which is exponential and so will be linear on a log plot. Similarly, $(1-{1\over 2})^s\leq (1-{1\over i})^s$, so $\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\geq \frac{1}{n}\sum_{i=2}^n \left(\frac{1}{2}\right)^s={n-1\over n}({1\over 2})^s$. Since $n$ is a constant, this is also just an exponential, so it must also be linear on a log plot. I believe that, since it is bounded between two functions that are of exponential order, that it also must be exponential order, so it also must be linear on a log plot.
This final assumption may be wrong, but I believe that it at least shows how one could model it well. Using desmos, I found the approximation $\exp(-\frac{s}{n}-\frac{\ln^{2} s}{\ln n})$, which is $O\Big(e^{s\over n}\Big)$ and only differs by about 12 for large $s$.
A: It suffices to analyze
$$
S_n=\sum_{k=2}^n\left(1-\frac1k\right)^s
$$
Using generalized binomial theorem, we have
$$
S_n=\sum_{k=2}^n\left\{1-\frac sk+\mathcal O_s\left(1\over k^2\right)\right\}
$$
Rearranging the terms, we get
$$
S_n=n-s\log n+\mathcal O_s(1)
$$
Consequently we have
$$
\frac1n\sum_{k=1}^n\left(1-\frac1k\right)^s=1-{s\log n\over n}+\mathcal O_s\left(\frac1n\right)
$$
