<s>[INST] 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.
 [/INST] 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:
$$
\lim_{n\to\infty} \sum_{k=1}^n n \int_{-\frac{1}{n}}^0 \left(- \frac{2x}{(1+\frac{k}{n})^3} \right)dx = \lim_{n\to\infty} \sum_{k=1}^n n \left(\frac{1}{n^2(1+\frac{k}{n})^3} \right)dx \\
= \int_0^1 \frac{dx}{(1+x)^3} = \frac{3}{8}.
$$
</s>