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.

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 $n \ge 1$.