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

Question

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.

Answer 1

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 $1/n\to 0$.

Answer 2

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:

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

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