Is the series $\sum_{n=1}^{\infty} \Bigl(1-\Bigl(1-\frac{1}{n^{1+\epsilon}}\Bigr)^n\Bigr)$ convergent? While i was solving a problem in probability theory I came across the following series 
$$\sum_{n=1}^{\infty} \biggl(1-\biggl(1-\frac{1}{n^{1+\epsilon}}\biggr)^n\biggr)$$
and in order to complete my solution I want to show the above series converges but I couldn't prove it (I still dont know if it converges).
I tried to go with Taylor expansion of $x \mapsto \ln(1-x^{1+\epsilon})$ but I couldn't get anything interesting.
Then I showed that $1-\bigl(1-\frac{1}{n^{1+\epsilon}}\bigr)^n \leq \frac{1}{n^\epsilon}$ but this doesn't help either.
Let me know if you have any idea!

You can get
explicit constants
rather than big or little oh
like this:
If $0 < x < 1$,
$-\ln(1-x)
=\sum_{k=1}^{\infty} \dfrac{x^k}{k}
\gt x$
and,
$\begin{array}\\
-\ln(1-x)
&=\sum_{k=1}^{\infty} \dfrac{x^k}{k}\\
&=x+\sum_{k=2}^{\infty} \dfrac{x^k}{k}\\
&\lt x+\sum_{k=2}^{\infty} \dfrac{x^k}{2}\\
&\lt x+\dfrac{x^2}{2(1-x)}\\
\end{array}
$
Therefore,
if $0 < x < \frac12$,
$-\ln(1-x)
\lt x+x^2
$
so
$-x-x^2
\lt\ln(1-x)
\lt -x
$.
Similarly,
if $0 < x < 1$,
$\exp(x)
=\sum_{k=0}^{\infty} \dfrac{x^k}{k!}
\gt 1+x$
and
$\exp(x)
\lt\sum_{k=0}^{\infty} x^k
=\dfrac1{1-x}
$.
Therefore,
for $0 < x < 1$,
$1-x \lt \exp(-x) \lt \dfrac1{1+x}$.
Therefore,
if
$n^{-c} < 1$,
then
$\begin{array}\\
(1-\frac1{n^{1+c}})^n
&=\exp(n\ln(1-\frac1{n^{1+c}}))\\
&\lt\exp(n(-\frac1{n^{1+c}}))\\
&=\exp(-n^{-c})\\
&\lt \dfrac1{1+n^{-c}}\\
\end{array}
$
so
$\begin{array}\\
1-(1-\frac1{n^{1+c}})^n
&\gt 1-\dfrac1{1+n^{-c}}\\
&=\dfrac{n^{-c}}{1+n^{-c}}\\
&=\frac12 n^{-c}\\
\end{array}
$
so the sum diverges
if $c \le 1$.
If $c > 1$ then
$\begin{array}\\
(1-\frac1{n^{1+c}})^n
&=\exp(n\ln(1-\frac1{n^{1+c}}))\\
&\gt\exp(n(-\frac1{n^{1+c}}-\frac1{n^{2+2c}}))\\
&=\exp(-n^{-c}-n^{-1-2c})\\
&\gt 1-(n^{-c}+n^{-1-2c})\\
\end{array}
$
so
$\begin{array}\\
1-(1-\frac1{n^{1+c}})^n
&\lt 1-(1-(n^{-c}+n^{-1-2c}))\\
&= n^{-c}+n^{-1-2c}\\
&\lt n^{-c}+n^{-3}\\
\end{array}
$
and the sum converges.
