I would like to find an equivalent of
$$ u_{n}-u_{\infty}=\sum_{k=1}^{n} \frac{n}{n^2+k^2}-u_{\infty} $$
Using Riemann sums, it is easy to show that:
$$ u_{n} \sim \frac{\pi}{4}=u_{\infty} $$
Using integrals, we have:
$$ \int_{1}^{n+1} \frac{n}{n^2+x^2} \mathrm dx \leq u_{n} \leq \int_{0}^{n} \frac{n}{n^2+x^2} \mathrm dx$$
$$ \arctan(1+1/n)-\arctan(1/n) \leq u_{n} \leq \frac{\pi}{4}$$
$$ \arctan(1+1/n)-\arctan(1/n)-\frac{\pi}{4} \leq u_{n} -\frac{\pi}{4}\leq 0$$
$$ \arctan(1+1/n)-\arctan(1/n)= \frac{\pi}{4}+\frac{1}{2n}-\frac{1}{n}+o(1/n)=\frac{\pi}{4}-\frac{1}{2n}+o(1/n) $$
So:
$$ -\frac{1}{2n}+o(1/n) \leq u_{n}-\frac{\pi}{4} \leq 0 $$
However the inequality prevents from writing $$ u_{n}-\frac{\pi}{4} \sim -\frac{1}{2n}$$ and numerical values seem to show that:
$$ u_{n}-\frac{\pi}{4} \sim -\frac{1}{4n}$$
Where did I go wrong?