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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?

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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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  • $\begingroup$ Sorry, but could you yet explain what this inequality comes from? $\endgroup$ – mrnobody Feb 7 '16 at 16:48
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    $\begingroup$ @scaremonger I added a highlighted section that gives a reference to an answer I posted in which I showed the inequality. - Mark $\endgroup$ – Mark Viola Feb 7 '16 at 16:56
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Firstly, you have a mistake in your last line: \begin{align}\lim_{n \rightarrow \infty} \int_{\mathbb{R}} I_{(0,n)(x)} \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx = \int_{0}^\infty \exp(-x) \, dx= 1 \end{align} and not $+\infty$ as you have it (because you integrated from $-\infty$ and not from $0$). You could also change variable first $y=\frac{x}{n}$: $$\int_{0}^{n} (1+ \frac{x}{n})^{n+1} \exp{(-2x)} dx=\int_{0}^{1} (1+y)^{n+1} \exp{(-2ny)} n\,dy$$ and now everything happens in $[0,1]$.

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    $\begingroup$ Oh, thanks. I've forgotten about the indicator... $\endgroup$ – mrnobody Feb 7 '16 at 16:36
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Seems easier to use dominated convergence. Hint: $1+t\le e^t$, so $(1+x/n)^{n+1}\le\dots$; for $n\ge 2$ this is less than or equal to $\dots$.

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  • $\begingroup$ $(e^{x})^{\frac{n+1}{n}}$? $\endgroup$ – mrnobody Feb 7 '16 at 16:32
  • $\begingroup$ @scaremonger $\lim_{n\to \infty}\frac{n+1}{n}=1$. $\endgroup$ – Mark Viola Feb 7 '16 at 16:45
  • $\begingroup$ I see, so, because the limit preserves inequalities, we obtain that $ (1+ \frac{x}{n})^{n+1} \le e^x$. $\endgroup$ – mrnobody Feb 7 '16 at 16:52
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    $\begingroup$ @scaremonger Yes, limits preserve inequalities. That does not show what you said! What you said is like saying $\lim 1/n <=0$, so $1/n\le 0$. Yes, the thing is less than $e^{n+1)/n$. And if $n\ge 2$ then $(n+1)/n\le3/2$. $\endgroup$ – David C. Ullrich Feb 7 '16 at 17:02

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