TITLE: Sums of arctangents
QUESTION [9 upvotes]: $$
\begin{align}
\arctan(x) = {} & \arctan(1) + \arctan\left(\frac{x-1} 2 \right) \\
& {} - \arctan\left(\frac{(x-1)^2} 4 \right) + \arctan\left(\frac{(x-1)^3} 8 \right) - \cdots
\end{align}
$$
Is this known?

REPLY [4 votes]: I've voted up Pietro Majer's incomplete answer and Michael Renardy's incomplete answer in the "comments" section.  Here's my own incomplete answer.
Here's how I got this series: start with the identity
$$
\arctan a - \arctan b = \arctan \frac{a-b}{1+ab}.
$$
From this we get
$$
\arctan x = \arctan 1 + \arctan\frac{x-1}{1+x}.
$$
Substituting 1 for $x$ everywhere in the last expression except the power of $x-1$, we get the 1st-degree term.  So we need to replace the last term above by the 1st-degree term plus another arctangent by using the basic identity above, and we get
$$
\arctan\frac{x-1}{1+x} = \arctan\frac{x-1}{2} + \arctan\frac{-(x-1)^2}{2(1+x) +(x-1)^2}.
$$
Then again substitute 1 for $x$ everwhere in the last term except in the power of $(x-1)$ in the numerator, to get the 2nd-degree term, and then write the last term above as the sum of the 2nd-degree term and another arctangent of a yet more complicated rational function.  And so on.
Does the sequence of arctangents of rational functions go to 0?  In some sense?  I don't know, nor do I know the general pattern.
I actually tried this first with $x-2$ instead of $x-1$; then I decided that $x-1$ already has enough initial unclarity.
I don't even know whether in some reasonable sense the process goes on forever.