I'm looking for a usable (something I can program) closed form or recursive formula for the coefficients $a_n$ of this series for an even-symmetry function of a real variable:
$$ \sum\limits_{n=0}^{\infty} a_n x^{2n} = \frac{x}{\arctan(x)} $$
for $\big|x\big| \le 1$ .
Ultimately, what I want is an expression like
$$ \arctan(x) = \frac{x}{\sum\limits_{n=0}^{\infty} a_n x^{2n}} $$
again, for $\big|x\big| \le 1$ .
I've approached this both from the POV of Taylor/Maclaurin series and tried to get a grip of using the Lagrange inversion theorem, but am not getting to an elegant place.
$$\begin{align} \sum\limits_{n=0}^{\infty} a_n x^{2n} &= \frac{x}{\arctan(x)} \\ \\ &= \frac{x}{\sum\limits_{k=0}^{\infty}\tfrac{(-1)^k}{2k+1} x^{2k+1}} \\ \\ &= \frac{1}{\sum\limits_{k=0}^{\infty}\tfrac{(-1)^k}{2k+1} x^{2k}} \\ \end{align}$$
or
$$ \left( \sum\limits_{n=0}^{\infty} a_n u^n \right) \left(\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1} u^k \right) = 1 $$
for $0 \le u \le 1$ .
I get this:
$$ \sum\limits_{n=0}^{\infty} a_n \sum\limits_{k=0}^{\infty} \frac{(-1)^k}{2k+1} u^{k+n} = 1 $$
and
$$ \sum\limits_{n=0}^{\infty} (-1)^n a_n \sum\limits_{k=n}^{\infty} \frac{(-1)^k}{2k-2n+1} u^{k} = 1 $$
I can see right away that $a_0 = 1$, but I should be able to yank out a recursion formula out of this. But I forgot how to do that. All of the net coefficients of the higher powers of $u^k$ should be $0$ for $k \ge 1$.
Denouement:
It's simpler than I thought.
So we have
$$ \sum\limits_{n=0}^{\infty} \sum\limits_{k=0}^{\infty} a_n \frac{(-1)^k}{2k+1} u^{k+n} = 1 $$
Let's let the exponent of $u$ be fixed to some integer $m \ge 0$. So $n+k=m$ and, for a given $n$, the only term in the inner summation that counts for the coefficient of $u^m$ is $k=m-n$.
We know for $m=0$ the summation is $1$.
$$ a_n \frac{(-1)^k}{2k+1} u^{k+n} \Big|_{n=k=0} = 1 $$
which results in $a_0=1$. For all other $m\ge 1$, the summation is
$$ \sum\limits_{n=0}^{m} a_n \frac{(-1)^{m-n}}{2(m-n)+1} u^m = 0 \qquad \forall u: 0 \le u \le 1$$
or
$$ \sum\limits_{n=0}^{m} a_n \frac{(-1)^{m-n}}{2(m-n)+1} = 0$$
Spinning off the last term for
$$ a_m + \sum\limits_{n=0}^{m-1} a_n \frac{(-1)^{m-n}}{2(m-n)+1} = 0 $$
or
$$ a_m = -\sum\limits_{n=0}^{m-1} a_n \frac{(-1)^{m-n}}{2(m-n)+1} $$
I confess that I didn't read the reference, I just needed to look at this more clearly. I guess with the additional substitution of $k=m-n$ we get:
$$ a_m = -\sum\limits_{k=1}^{m} a_{m-k} \frac{(-1)^k}{2k+1} $$
.
Here's one way of handling this. Note that
\begin{align} \sum\limits_{n=0}^{\infty} a_n x^{2n} \arctan(x) &= x \implies \\ \\ \left(\sum\limits_{n=0}^{\infty} a_n x^{2n}\right)\left( \sum\limits_{k=0}^{\infty}\tfrac{(-1)^k}{2k+1} x^{2k+1}\right) &= x \\ \left(\sum\limits_{n=0}^{\infty} a_n x^{2n}\right)\left( \sum\limits_{k=0}^{\infty}\tfrac{(-1)^k}{2k+1} x^{2k}\right) &= 1 \\ \end{align}
Then you can expand and equate coefficients:
\begin{align} a_0\cdot 1 &= 1 \implies a_0=1\\ (a_1-a_0/3)\cdot x^2 &= 0 \implies a_1=1/3\\ \vdots \end{align}
Another way, from here:
$$a_0 = 1$$
$$a_n = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{2k+1}a_{n-k}$$
Special thanks to Sangchul Lee for changing the indices nicely.