Infinite series for the arctangent from the tangent of half-angle formula From Hodge's biography of Turing:

He had found the infinite series for the "inverse tangent function", starting from the trigonometrical formula for $\tan\left(\frac{1}{2}x\right)$.*

The footnote states that he used no calculus and and also that "Perhaps the most remarkable thing was his seeing that such a series should exist at all."
So how do you get the infinite series for the inverse tangent from the half-angle formula for the tangent?
Thank you.
 A: Indeed using $2\arctan x = \arctan \dfrac{2x}{1-x^2}$ is the way to go. 
To simplify things, lets also assume that $\arctan x$ is odd, so $\arctan x = \displaystyle\sum_{n=0}^{\infty}a_nx^{2n+1}$. 
Then, using the Generalized Binomial Theorem, a change of indices, and a change of order of summation, he could have gotten: 
$\arctan \dfrac{2x}{1-x^2} = \displaystyle\sum_{k=0}^{\infty}a_k\left(\dfrac{2x}{1-x^2}\right)^{2k+1} = \displaystyle\sum_{k=0}^{\infty}a_k2^{2k+1}x^{2k+1}(1-x^2)^{-(2k+1)}$. 
$= \displaystyle\sum_{k=0}^{\infty}a_k2^{2k+1}x^{2k+1}\sum_{n = 0}^{\infty}\dbinom{2k+n}{2k}x^{2n} = \displaystyle\sum_{k=0}^{\infty}\sum_{n = 0}^{\infty}a_k2^{2k+1}\dbinom{2k+n}{2k}x^{2n+2k+1}$
$= \displaystyle\sum_{k=0}^{\infty}\sum_{n' = k}^{\infty}a_k2^{2k+1}\dbinom{n'+k}{2k}x^{2n'+1} = \sum_{n'=0}^{\infty}\left[\sum_{k = 0}^{n'}a_k2^{2k+1}\dbinom{n'+k}{2k}\right]x^{2n'+1}$. 
Hence, $2a_n = \displaystyle\sum_{k = 0}^{n}a_k2^{2k+1}\dbinom{n+k}{2k}$, i.e. $a_n = -\dfrac{1}{2^{2n+1}-2}\displaystyle\sum_{k = 0}^{n-1}2^{2k+1}\dbinom{n+k}{2k}a_k$ for $n \ge 1$. 
Since multiplying the series $a_n$ by any constant won't change the relationship $2\arctan x = \arctan \dfrac{2x}{1-x^2}$, Turing would have also needed the initial term $a_0 = 1$. 
From here, it is pretty easy to crank out the first few terms $a_1 = -\dfrac{1}{3}$, $a_2 = \dfrac{1}{5}$, $a_3 = -\dfrac{1}{7}$, etc. and guess that $a_n = \dfrac{(-1)^n}{2n+1}$. Although I'm not sure how he would have proven that $-\dfrac{1}{2^{2n+1}-2}\displaystyle\sum_{k = 0}^{n-1}2^{2k+1}\dbinom{n+k}{2k}\dfrac{(-1)^k}{2k+1} = \dfrac{(-1)^n}{2n+1}$ for all integers $n \ge 1$. 
A: If we say 
$$
\arctan x=\sum_{n=0}^\infty a_nx^n
$$
then  using $\tan x\approx x$ for small $x$, we get $a_0=0$ and $a_1=1$.
Now using @r9m's insight (see comment under question)
$$
2\arctan x=\arctan {2x\over 1-x^2}=\sum a_n2^nx^n(1+x^2+x^4+x^6+\cdots)^n
$$
For the $x^2$ term the factors are
$$
2a_2x^2=a_2(2x)^2
$$
thus $a_2=0$. Equating terms of $x^3$ gets
$$
2a_3x^3=a_3(2x)^3-2a_1x^3
$$
which gives us
$a_3=-1/3$. You can go on and on with this to get $a_4$ and $a_5$ etc. Turing may have noticed a pattern, gone further and worked out a recurrence relation.
