How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$ How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as
$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)\dfrac{x^{14}}{7}+...$
I have tried to use Cauchy product or long multiplication, but the result, although of course, are equivalent, is not as elegantly expressed as this expression. This is taken from Ferrar "A textbook of convergence" page 123.
For example, for $x^6$, I have $\dfrac{2}{5}-\dfrac{1}{9}$, while for $x^{10}$, I have $\dfrac{2}{9}+\dfrac{1}{25}-\dfrac{2}{21}$. I don't know how to make it pop out the partial Leibnitz series for the coefficients.
 A: Hint:
$f(x)=[\tan^{-1}(x)][\tanh^{-1}(x)]$
$f'(x)=\dfrac{ \tan^{-1}(x)}{1-x^2}+\dfrac{ \tanh^{-1}(x)}{1+x^2} $
If $|x|<1$:
$\displaystyle f'(x)=\left(\sum_{n=0}^{+\infty}  \dfrac{(-1)^n}{2n+1} x^{2n+1}   \right) \left(\sum_{n=0}^{+\infty}   x^{2n}  \right)   +  \left(\sum_{n=0}^{+\infty}  \dfrac{1}{2n+1} x^{2n+1}   \right) \left(\sum_{n=0}^{+\infty} (-1)^n  x^{2n}  \right)    $
Edit to show the idea
$f′(x)=\left(x−\dfrac{x^3}{3}+\dfrac{x^5}{5}−\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1+x^2+x^4+x^6+x^8...)+\left(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1−x^2+x^4−x^6+x^8...) $
By using Cauchy product:
$f′(x)=2x+ 2\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)x^5+2\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)x^9+...$
Then we integrate:
$f(x)=x^2+\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3} +\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5} +...  $
A: Since,
for $|x| < 1$,
$\arctan(x)
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{2n+1}
$,
so
$\arctan(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n}{2n+1}
$.
Similarly,
for $|x| < 1$,
$arctanh(x)
=\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1}
$,
so
$arctanh(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{x^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{x^n}{2n+1}
$.
Therefore
for $|x| < 1$,
$\arctan(\sqrt{x})arctanh(\sqrt{x})
=x\sum_{n=0}^{\infty}
x^n\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}
$
so
$\begin{array}\\
\arctan(x)arctanh(x)
&=x^2\sum_{n=0}^{\infty}
x^{2n}\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}\\
&=\sum_{n=0}^{\infty}
x^{2n+2}\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}\\
&=\sum_{n=1}^{\infty}
x^{2n}\sum_{k=0}^{n-1} \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-1-k)+1}\\
&=\sum_{n=1}^{\infty}
x^{2n}c_n\\
\end{array}
$
The coefficients are thus
$c_n
=\sum_{k=0}^{n-1} \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-1-k)+1}
$.
According to Wolfy,
the first few terms are
$x^2 + \dfrac{13 x^6}{45} + \dfrac{263 x^{10}}{1575} + 
\dfrac{36979x^{14}}{315315}+O(x^{18})
$.
It looks like
$c_{2n} = 0$
and
$c_{2n+1} \ne 0$.
The first should be
straightforward to prove
by reversing the order of summation.
A similar technique
should prove that
$c_{2n+1} > 0$.
As to a closed form
for the sum,
I don't know.
