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.
 A: 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 
$$u_n=\sum_{k=1}^nnA_k=\int_0^1\frac{1}{(1+x)^3}dx=\frac 38$$
A: Here's a rigorous derivation.
$$
\frac{n}{2} - \sum_{k=1}^n \frac{n^2}{(n+k)^2} = n \int_0^1 \frac{dx}{(1+x)^2} - \sum_{k=1}^n \frac{1}{(1+\frac{k}{n})^2} \\
= \sum_{k=1}^n \left(\int_{\frac{k-1}{n}}^{\frac{k}{n}} \frac{ n\, dx}{(1+x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\
= \sum_{k=1}^n \left(\int_{-\frac{1}{n}}^0 \frac{ n\, dx}{(1+\frac{k}{n} + x)^2} - \frac{1}{(1+\frac{k}{n})^2}\right) \\
= \sum_{k=1}^n \left(n\int_{-\frac{1}{n}}^0 \left(\frac{1}{(1+\frac{k}{n})^2} - \frac{2x}{(1+\frac{k}{n})^3} + \frac{3 x^2_*(x,n,k)}{(1+\frac{k}{n} +x_*(x,n,k))^4}\right)dx - \frac{1}{(1+\frac{k}{n})^2}\right)\\
= \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} + \frac{3 x^2_*(x,n,k)}{(1+\frac{k}{n}+x_*(x,n,k))^4}\right)dx.
$$
In the last two lines, we've used Taylor's theorem for $(1+\frac{k}{n}+x)^{-2}$. By that theorem with the Lagrange form of the remainder, $x_*(x,n,k)$ is some point in $(-\frac{1}{n},0)$. Given this bound on $x_*$, it is easy to see the term involving $x_*$ is $O(n^{-1})$, so we can ignore it in the limit $n\to\infty$. Continuing, we evaluate the limit of the remaining term:
$$
\lim_{n\to\infty} \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} \right)dx = \lim_{n\to\infty} \sum_{k=1}^n n \left(\frac{1}{n^2(1+\frac{k}{n})^3} \right)dx \\
= \int_0^1 \frac{dx}{(1+x)^3} = \frac{3}{8}.
$$
A: $$\dfrac1{(1+k/n)^2} = \left( 1 + \dfrac{k}n \right)^{-2} = 1 + \dfrac{(-2)}{1!} \dfrac{k}n + \dfrac{(-2)(-3)}{2!} \dfrac{k^2}{n^2} + \dfrac{(-2)(-3)(-4)}{3!} \dfrac{k^3}{n^3} + \cdots $$
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.
