Find:
lim
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}
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