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?