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.

$$\dfrac1{(1+k/n)^2} = \left( 1 + \dfrac{k}n \right)^{-2} = 1 + \dfrac{(-2)}{1!} \dfrac{k}n + \dfrac{(-2)(-3)}{2!} \dfrac{k^2}{n^2} + \dfrac{(-2)(-3)(-4)}{3!} \dfrac{k^3}{n^3} + \cdots $$
Now recall that
$$\sum_{k=1}^n \left(\dfrac{k}n \right)^m = \dfrac{n}{m+1} + \dfrac12 + \mathcal{O}(1/n)$$
Hence,
\begin{align}
\sum_{k=1}^n \dfrac1{(1+k/n)^2} & = n + \dfrac{(-2)}{1!} \left(\dfrac{n}2 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \dfrac{(-2)(-3)}{2!} \left(\dfrac{n}3 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \dfrac{(-2)(-3)(-4)}{3!} \left(\dfrac{n}4 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \dfrac{(-2)(-3)(-4)(-5)}{4!} \left(\dfrac{n}5 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \cdots 
\end{align}
The leading order term is $n-n+n-n+n-n + \cdots$, which when regularized gives us $\dfrac{n}2$. The next term is
$$\dfrac12 \left(-2+3-4+5-6 \pm \cdots\right)$$ which when regularized gives us $$\dfrac{-1+\eta(-1)}2 = \dfrac{-1+\dfrac14}2 = - \dfrac38$$
Hence,
$$\sum_{k=1}^n \left(1+\dfrac{k}n\right)^{-2} = \dfrac{n}2 - \dfrac38 + \mathcal{O}(1/n)$$
Hence, the limit is $\dfrac38$.
The argument, though it appears to be non-rigorous, can be made rigorous with careful regularization.
