Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that the limit of Riemann sum $\displaystyle \frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}$ is $\displaystyle \int_0^1 \frac{dx}{(1+x)^2}=\frac{1}{2}$ we find the indeterminate form $\infty\times 0$. How can we avoid this problem? Thanks for help.

Not sure if this is correct, but here goes anyway,
$$u_n=n\left(\int_0^1\frac{1}{(1+x)^2}dx-\frac 1n\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$$   
Let $$\begin{align} A_k 
&=\int_{\frac{k-1}{n}}^{\frac{k}{n}}\frac{1}{(1+x)^2}\mathrm{d}x -\frac{1}{n\left(1+\frac{k}{n}\right)^2} \\
& =\frac{1}{n\left(1+\frac{k}{n}\right)}\cdot\left(\frac{1}{1+\frac{k-1}{n}}-\frac{1}{1+\frac{k}{n}}\right) \\
&=\frac{1}{n^2}\cdot\frac{1}{\left(1+\frac kn\right)^2}\frac{1}{1+\frac{k-1}{n}} \end{align}$$
Then 
$$u_n=\sum_{k=1}^nnA_k=\int_0^1\frac{1}{(1+x)^3}dx=\frac 38$$
