Taylor expansion of $\;\dfrac{\operatorname{arctanh}(x)}{1+x^2}\,.$
Here is my trial:
Using the Taylor series of $\displaystyle\operatorname{arctanh}(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$ and $\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^n x^{2n}$
along with the Cauchy product:
$$\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right)=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k b_{n-k}\right)x^{n}$$
where $a_n=\frac{1}{2n+1}$ and $b_n=(-1)^n$, we have
$$\frac{\operatorname{arctanh}(x)}{1+x^2}=\sum_{n=0}^\infty (-1)^n\left(\sum_{k=0}^n \frac{(-1)^{k}}{2k+1}\right)x^{2n+1}$$
I know that $\displaystyle\sum_{k=0}^n \frac{1}{2k+1}=H_{2n}-\frac12H_n+\frac{1}{2n+1}$ but not sure about $\displaystyle\sum_{k=0}^n \frac{(-1)^{k}}{2k+1}$.
However, I feel it can be expressed in terms of harmonic numbers. Any idea?
Best Answer
I think that it is $$\frac{\operatorname{arctanh}(x)}{1+x^2}=\sum_{n=0}^\infty (-1)^{n\color{red}{+1}}\left(\sum_{k=0}^n \frac{(-1)^{k-1}}{2k+1}\right)x^{2n+1}$$ and the simplest form is $$\sum_{k=0}^n \frac{(-1)^{k-1}}{2k+1}=-\frac{(-1)^n}{2}\,\, \Phi \left(-1,1,\frac{2n+3}{2}\right)-\frac{\pi }{4}$$ and $$\Phi\left(-1,1,\frac{2n+3}{2}\right)=\frac{1}{2} \left(H_{\frac{2n+1}{4}}-H_{\frac{2n-1}{4}} \right)$$ Finally $$\color{blue}{\frac{\operatorname{arctanh}(x)}{1+x^2}=\frac 14 \sum_{n=0}^\infty \left(H_{\frac{2 n+1}{4}}-H_{\frac{2 n-1}{4} }+ (-1)^n\, \pi\right)\, x^{2n+1}}$$