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I'm trying to prove the following:

Let $f_{n}$ be a sequence defined by $f_{n}(x)=(1-(x/n))^n\ln(x)1_{[1,n]}(x)$ for every $x\in\mathbb{R}$ and for every $n\geq 1.$ Show that $$\displaystyle\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}f_{n}(x)dx=\int_{-\infty}^{\infty}e^{-x}\ln(x)1_{[0,\infty]}(x)dx.$$

My attempt is based in using dominated convergence because the limit of the sequence $f_{n}$ convergences without problem to $e^{-x}\ln(x)1_{[0,\infty]}(x).$

I'm stuck finding a function such as dominates the sequence $f_{n}$ and of course, it be integrable. I tried bounded $\ln(x)$ with identity function and $e^-x$; the problem here is that the sequence $(1-\frac{x}{n})^n$ is monotone decrecient.

Any kind of help is thanked in advance.

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There are two ways odf provung it.

  1. $\Bigl(1-\dfrac xn\Bigr)^n\chi_{[1,n]}(x)$ is increasing as a function of $n$ for all $x>0$, and the monotone convergence theorem.
  2. $\Bigl(1-\dfrac xn\Bigr)^n\chi_{[1,n]}(x)\le e^{-x}$ for all $x\ge0$ and the dominated convergence theorem.
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For $x>0$ you can use the inequality $\ln{(x)}\le x-1$, hence $$\frac{\ln(x)}{e^x}\le \frac{x-1}{e^x}$$ with $\int_{0}^{\infty}e^{-x}(x-1)dx=[-xe^{-x}]_{0}^{\infty}=0$.

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  • 1
    $\begingroup$ Thanks@Jimmy R. Your answer was great. I haven't tried to use that inequality. $\endgroup$ – Suiz96 Dec 1 '16 at 16:28

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