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 $(x-1)$, (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.
