For any real number $x$, $\tan^{-1}(x)$ denote unique real number $\theta$ in (-$\pi\over2$, $\pi\over2$) such that $\tan\theta=x$. Then evaluate
$\lim_{n\to\infty}$ $\sum_{m=1}^n \tan^{-1}$$1\over(1+m+m^2)$
How can I resolve that ? I don't know how to make to appear 1/n to evaluate in [0,1]
We have
$$\sum_{m = 1}^n \tan^{-1} \frac{1}{1 + m + m^2} = \sum_{m = 1}^n \tan^{-1} \frac{(m+1) - m}{1 + (m+1)m} = \sum_{m = 1}^n [\tan^{-1}(m+1) - \tan^{-1}(m)],$$
which is equal to
$$\tan^{-1}(n+1) - \tan^{-1}(1) = \tan^{-1}(n+1) - \frac{\pi}{4}$$
by telescoping. Taking the limit as $n\to \infty$ yields
$$\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.$$