Show $\lim_{n\to \infty} \int_n^\infty \left(1+\frac{x}{n} \right)^n e^{-x} \, dx=0.$ I want to show $$\displaystyle\lim_{n\to \infty} \int_n^\infty \left(1+\frac{x}{n} \right)^n e^{-x} \, dx=0.$$
Hint is given ;
HINT :
$\displaystyle\lim_{n\to \infty} u_n=0,$ where $u_n=\displaystyle\max_{x\geqq n}\left(1+\frac{x}{n}\right)^n e^{-x}.$

So far, I tried to do simply by $\epsilon-n.$
Let $\epsilon >0.$
Since $\lim u_n=0,$ there is $N\in \mathbb N$ s.t. $n\geqq N \Rightarrow u_n<\epsilon.$
When $n\geqq N$, $\left| \int_n^\infty \left(1+\frac{x}{n}\right)^n e^{-x} \, dx \right| \leqq \int_n^\infty \left| \left(1+\frac{x}{n}\right)^n e^{-x} \right| \, dx =\int_n^\infty \left(1+\frac{x}{n}\right)^n e^{-x}  \, dx \leqq \int_n^\infty  u_n \, dx < \int_n^\infty \epsilon \, dx.$
This fails because the upper limit of integral is $\infty.$ I don't know how I should handle the upper limit $\infty.$
I'd like you give me any idea.
 A: For $x \geq 1$, we have
$$ \log(1+x) \leq x - \frac{x^2}{2(1+x)} \leq \frac{3}{4}x $$
(The first inequality actually holds for all of $x \geq 0$ and can be proved by differentiation.) Using this, we get
$$ \left(1 + \frac{x}{n}\right)^n e^{-x} \mathbf{1}_{[n,\infty)}(x) \leq e^{-x/4}. $$
So by the dominated convergence theorem,
$$ \lim_{n\to\infty} \int_{n}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \, \mathrm{d}x
= \int_{0}^{\infty} \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \mathbf{1}_{[n,\infty)}(x) \, \mathrm{d}x
= 0. $$
A: For $x\geqslant0$, the function $x\mapsto(1+x/n)^ne^{-x/2}$ attains the maximum at $x=n$ (easy to check; the logarithmic derivative is $1/(1+x/n)-1/2$...). Thus $(1+x/n)^ne^{-x/2}\leqslant2^ne^{-n/2}$ and $$0\leqslant\int_n^\infty(1+x/n)^ne^{-x}\,dx\leqslant2^ne^{-n/2}\int_n^\infty e^{-x/2}\,dx=2(2/e)^n\underset{n\to\infty}{\longrightarrow}0.$$
A: Since the map $t\mapsto\Big(1+\frac{1}{t}\Big)^t$ is increasing in $t>0$ (see here), and $f_m(x)=\Big(1+\frac{x}{m}\Big)^me^{-x}$ is decreasing in $x>0$ for all $m$ ($f'_m(x)=-x\big(1+\frac{x}{m}\big)^{m-1}e^{-x}<0$), we have that
$$\begin{align}\mathbb{1}_{(n,\infty)}(x)\Big(1+\frac{x}{n}\Big)^ne^{-x}&=\sum^\infty_{m=n}\mathbb{1}_{(m,m+1]}(x)\Big(1+\frac{x}{n}\Big)^ne^{-x}\\
&\leq \sum^\infty_{m=n}\mathbb{1}_{(m,m+1]}(x)\Big(1+\frac{x}{m}\Big)^me^{-x}\\
&\leq\sum^\infty_{m=n}(2/e)^m\mathbb{1}_{(m,m+1]}(x)
\end{align}$$
Integration yields
$$\int^\infty_nf_n(x)\,dx\leq\sum^\infty_{m=n}(2/e)^m=(2/e)^n\frac{1}{1-\tfrac2e}\xrightarrow{n\rightarrow\infty}0$$
