Evaluate $\lim_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^{2n}\frac{1-\ln(1+\frac{k}{n})}{(1+\frac{k}{n})^2}$

As shown in the title, I'm evaluating the following:$$\lim_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^{2n}\frac{1-\ln(1+\frac{k}{n})}{(1+\frac{k}{n})^2}$$ And I get stuck. Any ideas are welcome.

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The last substitution doesn't make sence. Previous exrcise was given to simplify evaluation of integrals you will face in while calulating this sum – Norbert Nov 14 '12 at 14:36
Think about Riemann sums – Pedro Tamaroff Nov 14 '12 at 14:36
Have you tried expanding $log(1+x) = x - x^2/2 + \ldots$? I think this can be useful, because the $k/n < 1$. – m0nhawk Nov 14 '12 at 14:40
It turns into sum of infinite sum, even troublesome. – Michael Li Nov 14 '12 at 14:57
@m0nhawk That will overcomplicate things. – Pedro Tamaroff Nov 14 '12 at 15:11

Consider funstion $$f(x)=\frac{1-\log (1+x)}{(1+x)^2}$$ and partition $\{k/n:k=\overline{1,2n}\}$ of the interval $[0,2]$. Then Riemann sum for given partition will be... Then recall Riemann sums tends to integral when partition becomes smaller, i.e. when $1/n\to 0$.

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Consider the partitions

$$\left\{0<\frac{1}{n}<\frac{2}{n}<\ldots <\frac{2n}{n}=2\right\}\,\,\text{of the interval}\,[0,2]\,\,,\,\,n\in\Bbb N$$

Choosing the right-end points of each interval, we get:

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1-\log\left(1+\frac{k}{n}\right)}{\left(1+\frac{k}{n}\right)^2}=\int_0^2\frac{1-\log(1+x)}{(1+x)^2}\,dx=$$

$$=\left.-\frac{\log(x+1)+1}{x+1}\right|_0^2=\ldots$$

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The symbol between the limit and the integral must be an "=". Unless that, everything is fine. – leo Nov 14 '12 at 16:19
Of course, thanks – DonAntonio Nov 14 '12 at 16:22
The last step is weird, how can you reach it? – Michael Li Nov 18 '12 at 3:11
What do you mean "weird"? That's the integral's primitive, of course: putting $$u=\log x\;\;,\;u'=\frac{1}{x}\;;\;v'=\frac{1}{x^2}\;\;,\;v=-\frac{1}{x}$$one gets at once integrating by parts: $$\int\frac{\log x\,dx}{x^2}=-\frac{\log x+1}{x}$$Now pass from $\,x\,$ to $\,x+1\,$...voila! – DonAntonio Nov 18 '12 at 3:48
Yes, of course...but you know $$\int\frac{1}{x^2}dx=-\frac{1}{x}$$right? – DonAntonio Nov 18 '12 at 8:32