Convergence Proof Using Dominated Convergence Theorem I am trying to prove that the 
$A_n=n\int_{1}^{+\infty}\left(1-e^{-\frac{1}{x-1}}\right) e^{-x}\left(1-e^{-x}\right)^{n-1}$ $\Rightarrow$
$\lim_{n\to\infty}A_n=0$ .
In fact, by defining $f_n(x)=\color{red}{n}\left(1-e^{-\frac{1}{x-1}}\right) e^{-x}\left(1-e^{-x}\right)^{n-1}$, it is straightforward to show that $f_n$ converges pointwise to zero function as n tends to infinity. So, if I want to use Dominated Convergence Theorem, I should show that there exists an integrable function $g(x)$ such that $|f_n(x)|\leq g(x)$ for all $n$. 
My problem is I cannot find $g(x)$ to satisfy the constraints. 
Thank you for your help and comments.
 A: As you were discovering, your problem is not really a DCT situation. Perhaps you have seen the following result, or something like it: Let $0<b\le 1.$  Suppose $f$ is continuous on $[0,b]$ with $f(0) = 0.$ Then
$$\lim_{n\to \infty} n\int_0^b f(x)(1-x)^{n-1}\,dx = 0.$$
Now in the given problem, make the substitution $x= \ln (1/y).$ Your expression turns into
$$n\int_0^{1/e} f(y)(1-y)^n\,dy,$$
with $f$ satisfying the hypotheses above. The result follows.
A: Splitting the integral might be a good start. 
Given $\epsilon>0$, write 
$$
A_n=\int_1^{\infty} =  \int_{1}^{x_n} + \int_{x_n}^{\infty}= C_n+D_n.$$
The sequence $x_n$ will be determined later so that $x_n\rightarrow\infty$ as $n\rightarrow\infty$. 
We have by substitution $t=1-e^{-x}$, $dt=e^{-x}dx$, 
$$\begin{align}
C_n&\leq n\int_{1}^{x_n} e^{-x} (1-e^{-x})^{n-1} dx = n \int_{1-e^{-1}}^{1-e^{-x_n}} t^{n-1} dt \\ &\leq (1-e^{-x_n})^n     \leq   \exp(-\frac{n }{e^{x_n}}),\end{align}
$$
and for sufficiently large $n$, we have
$$\begin{align}
D_n&\leq n\int_{x_n}^{\infty} (1-e^{-\frac1{x_n-1}}) e^{-x}(1-e^{-x})^{n-1} dx \\ &\leq  \frac n{x_n-1}\int_{x_n}^{\infty} e^{-x}(1-e^{-x})^{n-1}dx \leq \frac n{x_n-1}\cdot \frac1n=\frac1{x_n-1}.\end{align}$$
Take $e^{x_n} = \sqrt n$.  Then 
$$
\lim_{n\rightarrow\infty} C_n = 0, \ \ \lim_{n\rightarrow\infty} D_n = 0.$$
Therefore, $\lim_{n\rightarrow\infty} A_n = 0$. 
