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. 1 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$(x-1)$, 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.