How do I find the infinite sum of $\arctan(2^{-n})$?

Question

I am looking at the CORDIC Algorithm and I want to show that iterative rotations by the nth angle given by arctan $2^{-n}$ is greater than $\frac{\pi}{2}$.

$$ \sum_{n=0}^{\infty} \arctan 2^{-n} $$

Answer 1

$$\sum_{n=0}^\infty \tan ^{-1}\left(\frac{1}{2^n}\right)=\frac \pi 4+\sum_{n=1}^\infty \tan ^{-1}\left(\frac{1}{2^n}\right)$$ Using Taylor expansion $$\tan^{-1}( t) = \sum _{p=0}^{\infty } (-1)^p\frac{t^{2 p+1}}{2p+1} \qquad \forall t \in~ ]-1,1[,$$ and Fubini's theorem, one gets $$\sum_{n=1}^\infty \tan ^{-1}\left(\frac{1}{2^n}\right)=\sum _{p=0}^{\infty } \frac{(-1)^p}{\left(2^{2 p+1}-1\right) (2 p+1)}$$ which converge very fast since $$a_p=\frac{1}{\left(2^{2 p+1}-1\right) (2 p+1)} \quad\implies \quad\underset{p\to \infty }{\text{limit}}\left(\frac{a_{p+1}}{a_p} \right)=\frac 14.$$ The series $\sum_p (-1)^p/((2p+1)(2^{2p+1}-1))$ fulfills the conditions of the alternating series test, so the partial sums $$1,\frac{20}{21},\frac{3121}{3255},\frac{395902}{413385},\frac{123888 89}{12933045},\frac{278947680568}{291213374265},\cdots.$$ are larger than $\frac{20}{21} >\frac \pi 4$ and then the full summation larger than $\frac \pi 2$.

Edit

Faster would be to use $$\tan^{-1}( t) > \frac{3 t}{3+t^2} \qquad \forall t >0$$ $$3\sum_{n=1}^\infty \frac{ 2^{-n}}{3+2^{-2 n}}=$$ $$\frac{\sqrt{3} \pi }{2 \log (2)}+i\,\frac{ \sqrt{3}}{2 \log (2)}\left(\psi _2^{(0)}\left(\frac{\log (12)+i \pi}{2log (2)}\right)-\psi _2^{(0)}\left(\frac{\log (12)-i \pi}{2\log (2)}\right)\right)$$ which is $0.955696 > \frac \pi 4$.

Answer 2

By the strict concavity of $\tan^{-1}(x)$, for $0\lt x\lt1$ $$ \tan^{-1}(x)\gt\frac\pi4x\tag1 $$ Thus, $$ \begin{align} \sum_{k=0}^\infty\tan^{-1}\left(\frac1{2^k}\right) &=\frac\pi4+\sum_{k=1}^\infty\tan^{-1}\left(\frac1{2^k}\right)\tag{2a}\\ &\gt\frac\pi4+\frac\pi4\sum_{k=1}^\infty\frac1{2^k}\tag{2b}\\ &=\frac\pi2\tag{2c} \end{align} $$