integration - Find $lim_{n rightarrow infty} int_0^n (1+ frac{x}{n})^{n+1} exp(-2x) , dx$

Find:

$$\lim_{n \rightarrow \infty} \int_0^n \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx$$

The sequence $\left(1+ \frac{x}{n}\right)^{n+1} \exp{(-2x)}$ converges pointwise to $\exp{(-x)}$. So if we could apply Lebesgue's monotone convergence theorem, we have:

\begin{align}
\lim_{n \rightarrow \infty} \int_0^n \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx &=\lim_{n \rightarrow \infty} \int_{\mathbb{R}} I_{(0,n)(x)} \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx \\[10pt]
&= {} \int_{- \infty}^\infty \exp(-x) \, dx= +\infty

\end{align}

Can someone help me to prove that the sequence of integrands is monotone?

Answer

Using the indicator function, we can write

$$\int_0^n \left(1+\frac xn\right)^{n+1}e^{-2x}\,dx=\int_0^\infty \left(1+\frac xn\right)^{n+1}e^{-2x}\xi_{[0,n]}(x)\,dx$$

Then, note that for $0\le x\le n$, we have

$$\left(1+\frac xn\right)^{n+1}=\left(1+\frac xn\right)\left(1+\frac xn\right)^{n}\le 2e^{x}$$

EDIT: The OP requested a source for the preceding inequality.
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the exponential function satisfies the inequality $$e^x\ge \left(1+\frac xn\right)^{n}$$for $x>-n$.

We can assert, therefore, that

$$\left(1+\frac xn\right)^{n+1}e^{-2x}\xi_{[0,n]}(x)\le 2e^{-x}$$

Applying the Dominated convergence theorem, we have

$$\begin{align}
\lim_{n\to \infty} \int_0^n \left(1+\frac xn\right)^{n+1}e^{-2x}\,dx&=\int_0^\infty \lim_{n\to \infty}\left(\left(1+\frac xn\right)^{n+1}\xi_{[0,n]}(x)\right)\,e^{-2x}\,dx\\\\
&=\int_0^\infty e^{x}\,(1)\,e^{-2x}\,dx\\\\
&=1
\end{align}$$