TITLE: How to evaluate $\sum_{n=1}^m 2^n \arctan 2^n \theta$
QUESTION [6 upvotes]: I need to evaluate $$\sum_{n=1}^m 2^n \arctan 2^n \theta$$ as a function of $m$ and $\theta$. All I've done so far is write out the series explicitly:
$$\sum_{n=1}^m 2^n \arctan 2^n \theta = 2 \arctan 2\theta + 4\arctan 4\theta + 8\arctan 8\theta + \cdots + 2^m \arctan 2^m \theta$$ 
and I initially considered pairing every two terms up to use the $\arctan x + \arctan y$ trick, but it doesn't work because each $\arctan$ term has a different coefficient.

REPLY [6 votes]: Not for nothing, but just in case the OP really wants to evaluate
$$\sum_{n=1}^m 2^n \tan{2^n \theta} $$
then use the fact that
$$\cot{x}-2 \cot{2 x} = \tan{x} $$
and let $x=2^n \theta$. In this case, we get a telescoping sum with the result
$$\sum_{n=1}^m 2^n \tan{2^n \theta} = 2\cot{2 x} - 2^{m+1} \cot{2^{m+1} \theta}$$