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.
Best Answer
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 $$0\leqslant\int_n^\infty(1+x/n)^ne^{-x}\,dx\leqslant2^ne^{-n/2}\int_n^\infty e^{-x/2}\,dx=2(2/e)^n\underset{n\to\infty}{\longrightarrow}0.$$