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I'd like to find

$\lim_{n \to \infty} \frac{1}{n}\int_{0}^{n}\frac{x\ln(1+\frac{x}{n})}{1+x}dx$

I am wondering if it is true that it converges to $\ln2$?

thanks in advance

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    Without doing work I would try to see if the Dominated Convergence Theorem applies. – 2017-02-27
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    If you want a real analysis answer, see the duplicate here http://math.stackexchange.com/a/2159980/269764 – 2017-02-27
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    @Dear Brevan Ellefsen, I appreciated it I find your answer very useful also I am sorry, I sent this question – 2017-02-27

2 Answers 2

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The involved integral equals $$ \int_{0}^{1}\log(1+x)\frac{nx}{nx+1}\,dx $$ that by the dominated convergence theorem converges to $$ \int_{0}^{1}\log(1+x)\,dx = \color{red}{\log\left(\frac{4}{e}\right)} $$ as $n\to +\infty$.

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    @dear Jack D'Aurizio thanks you answer, would you mind explaining your answer a little more – 2017-02-27
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    Given that the tag is real analysis, dominated convergence theorem is a bit powerful – 2017-02-27
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    @user62498: the first step is to perform the substitution $x\to nx$, the second step is to apply the dominated convergence theorem. The sequence of functions $f_n(x)=\frac{nx}{1+nx}$ is pointwise convergent to $1$ on the interval $(0,1)$. – 2017-02-27
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    @BrevanEllefsen: true, but this is the most simple answer. Otherwise we may apply squeezing: the idea stays pretty the same, the proof gets longer. – 2017-02-27
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    While the DCT *is* very powerful, it also has the advantage of... working, and being quite systematic. If one knows it, refraining from using it in favor of ad hoc, non-generalizable, and longer arguments, just for the sake of it, is a tad odd. (Now, if it's not known, or explicitly forbidden to rely on the DCT for an exercise, sure.) – 2017-02-27
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From L'Hospital's Rule we have

$$\begin{align} \lim_{n\to \infty}\frac1n\int_0^n\frac{x\log\left(1+\frac xn\right)}{1+x}\,dx&=\lim_{n\to \infty}\left(\frac{n\log(2)}{n+1}-\frac1n\int_0^n \frac{x^2}{(1+x)(n+x)}\right)\\\\ &=\log(2)-\lim_{n\to \infty}\frac1{n(n-1)}\int_0^n\left(\frac{nx}{n+x}-\frac{x}{1+x}\right)\,dx\\\\ &=\log(2)-\lim_{n\to \infty}\frac{1}{n(n-1)}\left.\left((n-1)x-n^2\log(x+n)+\log(x+1)\right)\right|_{x=0}^{x=n}\\\\ &=\log(2)-\lim_{n\to \infty}\frac{n(n-1)-n^2\log(2)+\log(n+1)}{n(n-1)}\\\\ &=2\log(2)-1 \end{align}$$

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    Small typo. The first limit is $\lim_{n\to \infty}$. – 2017-02-27