I need some help calculating the following integral in measure theory:
$$\int_{0}^{n} (1-\frac{x}{n})^n e^{\frac{x}{2}}dx$$
So far what I tried was the following:
$$\int_{0}^{n} (1-\frac{x}{n})^n e^{\frac{x}{2}}dx$$
$$= \int_{\mathbb{R}_+} (1-\frac{x}{n})^n e^{\frac{x}{2}} \cdot \mathbb{1}_{x \leq n} \lambda (dx)$$
Then I calculated the limit of the term in the integral:
$$\lim_{n \to \infty} (1-\frac{x}{n})^n e^{\frac{x}{2}} \cdot \mathbb{1}_{x \leq n} = e^{x}e^{\frac{x}{2}} = e^{-\frac{x}{2}}$$
And:
$$\int_{0}^{+ \infty} e^{-\frac{x}{2}}dx = 2$$
Now I wanted to use either the dominated convergence theorem or the monotone convergence theorem, but for either theorem I need either to show that $(1-\frac{x}{n})^n e^{\frac{x}{2}}$ converges monotonically or that it's upper bounded by $e^{-\frac{x}{2}}$. I tried different approaches, but none of them helped me. Has anyone an idea?
If $0\le x\le n$, then $$ \Bigl(1-\frac{x}{n}\Bigr)^n\le e^{-x}. $$ This is equivalent to $$ \log\Bigl(1-\frac{x}{n}\Bigr)\le-\frac{x}{n}, $$ which follows fron the inequality $\log(1-x)\le-x$ if $0\le x<1$.