# Solving an infinite series containing $\arctan$

I need to compute:

$$\tan\bigg(\arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{2}{9}\right)+ \arctan\left(\frac{1}{8}\right)+\arctan\left(\frac{2}{25}\right)+\arctan\left(\frac{1}{18}\right)+\ldots\bigg)$$

I proceeded as follows. The series is $\left(\sqrt{2}/(n+1)\right)^2$, so the sum inside $\tan$ can be simplified as:

$S=\sum_{n=1}^\infty \arctan(n+2)-\arctan(n)$. How do I proceed from here?

• Notice $$\tan^{-1}(n+2) - \tan^{-1}(n) = \tan^{-1}\frac{1}{n} - \tan^{-1}\frac{1}{n+2}\\ = \left( \tan^{-1}\frac{1}{n} + \tan^{-1}\frac{1}{n+1} \right) - \left( \tan^{-1}\frac{1}{n+1} + \tan^{-1}\frac{1}{n+2}\right)$$ The series inside the outermost $\tan(\cdot)$ is a telescoping one! – achille hui Jul 2 '15 at 13:05

You've already done the hard part. This is a telescoping series. As mathlove has shown, the $m$-th partial sum is $\arctan\left(m+1\right)+\arctan\left(m+2\right)−(\arctan1+\arctan2)$. The first two terms both go to $\pi/2$ as $m\to\infty$, so they provide a contribution $\pi$ that doesn't change the value of the outer tangent. Thus the result is $-\tan\left(\arctan1+\arctan2\right)$, and another application of the addition theorem that you already applied yields $3$.
Note that $$\sum_{n=1}^{m}(\arctan(n+2)-\arctan(n))$$$$=(\arctan(3)-\arctan(1))+(\arctan(4)-\arctan(2))+(\arctan(5)-\arctan(3))+\cdots+(\arctan(m)-\arctan(m-2))+(\arctan(m+1)-\arctan(m-1))+(\arctan(m+2)-\arctan(m))$$ $$=-\arctan(1)-\arctan(2)+\arctan(m+1)+\arctan(m+2)$$ $$=(\arctan(m+2)-\arctan(2))+(\arctan(m+1)-\arctan(1))$$ $$=\arctan\frac{m+2-2}{1+2(m+2)}+\arctan\frac{m+1-1}{1+1\cdot(m+1)}$$ $$=\arctan\frac{1}{2+\frac 5m}+\arctan\frac{1}{1+\frac 2m}$$
Hence, we have $$S=\lim_{m\to\infty}\left(\arctan\frac{1}{2+\frac 5m}+\arctan\frac{1}{1+\frac 2m}\right)=\arctan\frac 12+\arctan 1.$$
Then, the answer will be $$\tan\left(\arctan\frac 12+\arctan 1\right)=\frac{\frac 12+1}{1-\frac 12\cdot 1}=\color{red}{3}.$$