Lebesgue Integral computation Compute $\lim\limits_{n \to \infty} \int_0^n (1 - \frac{x}{n})^n e^{x/2} dx $, justify your calculation
My Try: Use Lebesgue Dominance Convergence Theorem $f_n = (1 -\frac{x}{n})^n e^{x/2}$ and $|f_n| \leq g$ where $g = e^{-x} e^{x/2}$. If i do this i have to prove that $(1 -\frac{x}{n})^n <e^{-x}$ and i don't know how to prove it, any idea, any help, Thanks
 A: We can show that $\left(1-\frac xn\right)^n$ monotonically increases by straightforward application of Bernoulli's Inequality.
Let $a_n=\left(1-\frac xn\right)^n$.  Then, we have
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\left(\frac{n(n+1-x)}{(n+1)(n-x)}\right)^{n+1}\left(1-\frac xn\right)\\\\
&=\left(1+\frac{x}{(n+1)(n-x)}\right)^{n+1}\left(1-\frac xn\right) \tag 1\\\\
&\ge\left(1+\frac{x}{(n-x)}\right)\left(1-\frac xn\right) \tag 2\\\\
&=1
\end{align}$$
for $x\ne n$, where in going from $(1)$ to $(2)$, we used Bernoulli's Inequality.  And for $x=n$, $a_n=0<a_{n+1}$ and therefore, $a_n$ monotonically increases to $e^{-x}$ for all $0\le x\le n$.
Therefore, we have for $x\ge 0$, 
$$\left|\left(1-\frac xn\right)^ne^{x/2}\right|\le e^{-x/2}$$
Therefore, we can write
$$\begin{align}
\left|\int_0^n \left(1-\frac xn\right)^ne^{x/2}\,dx\right|&=\left|\int_0^{\infty} \left(1-\frac xn\right)^ne^{x/2}\,\chi_{x\in[0,n]}\,dx\right|\\\\
&\le \int_0^{\infty} \left|\left(1-\frac xn\right)^n\right|\,e^{x/2}\,dx\\\\
&\le \int_0^{\infty}e^{-x/2}\,dx\\\\
&=2
\end{align}$$
whereupon invoking the Monontone Convergence Theorem reveals 
$$\lim_{n\to \infty}\int_0^n \left(1-\frac xn\right)^ne^{x/2}\,dx=2$$
