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I am trying to find the following limit. Let $X = [0,\infty)$ and $\mathbb B$ denote the Borel subsets in $[0,\infty)$, $\lambda$ the Lebesgue measure. Let $f_n : [0, \infty) \to \mathbb R$ be given by \begin{align*} f_n(x) = \left(\frac{n+x}{n+2x}\right)^n. \end{align*} Find $\lim_{n \to \infty} \int_{[0,n]} f_n d\lambda$.

I wanted to do this with Dominated Convergence, but I failed to find a function which dominates and is moreover integrable on $[0,\infty)$. ($g(x) = 1$ doesn't work.)

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If I make no mistakes then $m:=n+2x$ gives $\left(\frac{n+x}{n+2x}\right)^{n}=\left(1-\frac{x}{m}\right)^{m}e^{-2x\ln\left(‌​1-\frac{x}{m}\right)}\rightarrow e^{-x}$. Can that help? –  drhab Jan 17 at 12:56

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We have that \begin{align} f_n(x) = \left(\frac{n+x}{n+2x}\right)^n=\left(1-\frac{x}{n+2x}\right)^n\le \mathrm{e}^{-x} \end{align} for $x\in [0,n]$, and thus Lebesgue Dominated Convergence Theorem is applicable, i.e., $$ \lim_{n\to\infty}\int_0^n \left(\frac{n+x}{n+2x}\right)^n dx=\int_0^\infty\left(\lim_{n\to\infty} \left(\frac{n+x}{n+2x}\right)^n \right)\,dx=\int_0^\infty \mathrm{e}^{-x}\,dx=1. $$

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Use a change of variables $t=x/n$ to obtain $$n \int_0^1 \Bigl(\frac{1+t}{1+2t}\Bigr)^n dt \leq 2n \int_0^1 \Bigl( \frac{1+t}{1+2t}\Bigr)^{n-1} \frac{1}{1+2t} dt = -2 \Bigl( \frac{1+t}{1+2t}\Bigr)^n\Bigr|_0^1 \leq 2. $$

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But $\frac{d}{dt} \left(-2 \left(\frac{1+t}{1+2t}\right)^n\right) = 2n \left(\frac{1+t}{1+2t}\right)^{n-1} \cdot \frac{1}{(1+2t)^2}$ –  numerion Jan 17 at 15:16
    
Correct, so you can lob off another $1/(1+2t)$ and have the upper bound as $(\cdot)^{n-1}$, using the fact that $n/(n-1)$ is bounded. –  JPi Jan 17 at 15:30
    
btw: my derivation essentially establishes uniform integrability. The value of the integral that you're after is one. –  JPi Jan 17 at 15:31

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