Taylor expansion of $\frac{\operatorname{arctanh}(x)}{1+x^2}$

Question

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?

Answer 1

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 $$\color{blue}{\frac{\operatorname{arctanh}(x)}{1+x^2}=\frac 14 \sum_{n=0}^\infty \left(H_{\frac{2 n+1}{4}}-H_{\frac{2 n-1}{4} }+ (-1)^n\, \pi\right)\, x^{2n+1}}$$

Answer 2

A solution by Cornel Ioan Valean (in the sense of establishing a simple relation with harmonic numbers and digamma function)

A short introduction

By analogy with the harmonic number, $\displaystyle H_n=\sum_{k=1}^n \frac{1}{k}$, and the skew-harmonic number, $\displaystyle \overline{H}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{k}$, we may have the notations $\displaystyle O_n=\sum_{k=1}^n \frac{1}{2k-1}$ and $\displaystyle \overline{O}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1}$, where the latter may be related to the sum we are interested in, that is $\displaystyle \sum_{k=0}^n (-1)^{k-1}\frac{1}{2k+1}$. During the work, I'll use the variant $\overline{O}_n$ and we easily note that $\displaystyle \sum_{k=0}^n (-1)^{k-1}\frac{1}{2k+1}=-\overline{O}_n+(-1)^{n-1}\frac{1}{2n+1}$.

Before proceeding further, we might find helpul to take a look into the well-known and less sophisticated cousin of $\displaystyle \overline{O}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1}$, that is, $\displaystyle \overline{H}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{k}$. The skew-harmonic number, or even the more general case, that is, the generalized skew-harmonic numbers defined by $\displaystyle \overline{H}_n^{(m)}=\sum_{k=1}^n (-1)^{k-1}\frac{1}{k^m}$ may expressed in terms of Digamma and harmonic numbers. Such relations we might often find extremely useful! For example, in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), Sect. 4.18, pages 417-418, we have the folowing relations:

$$i) \ \psi\left(\frac{n}{2}\right)$$ $$=-\log(2)-\gamma-\log(2)(-1)^{n-1}-2\frac{1}{n}+H_n+(-1)^{n-1} \overline{H}_n;$$ $$ii) \ \psi\left(\frac{n}{2}+1\right)$$ $$=-\log(2)-\gamma-\log(2)(-1)^{n-1}+H_n+(-1)^{n-1} \overline{H}_n;$$ $$ iii) \ \psi^{(m)}\left(\frac{n}{2}\right)$$ $$=(-1)^{m-1} 2^{m+1} m!\frac{1}{n^{m+1}}+ (-1)^{m+n}(2^m-1) m!\zeta (m+1)$$ $$+(-1)^{m-1} 2^m m! \left(\zeta (m+1)-H_n^{(m+1)}\right)-(-1)^{m+n}2^m m! \overline{H}_n^{(m+1)};$$ $$iv)\ \psi^{(m)}\left(\frac{n}{2}+1\right)$$ $$= (-1)^{m+n}(2^m-1) m!\zeta (m+1)+(-1)^{m-1} 2^m m! \left(\zeta (m+1)-H_n^{(m+1)}\right)$$ $$-(-1)^{m+n}2^m m! \overline{H}_n^{(m+1)},$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1$, is the $n$th generalized harmonic number of order $m$, $\overline{H}_n^{(m)}=1-\frac{1}{2^m}+\cdots+(-1)^{n-1}\frac{1}{n^m}, \ m\ge1$, represents the $n$th generalized skew-harmonic number of order $m$, $\zeta$ denotes the Riemann zeta function, and $\psi^{(n)}$ designates the Polygamma function.

Are such relations useful? Absolutely! Here is an example where the work is performed with polygamma function and after getting the closed form in terms of polygamma, we are able to recover the answer in terms of the generalized skew-harmonic numbers: $$\int_0^1 x^{n-1} \log^3(x)\operatorname{Li}_2\left(\frac{1-x}{2}\right) \textrm{d}x$$ $$=\left(\frac{3}{8}\pi^2\zeta(3)+\frac{7}{120}\log(2)\pi^4-12\zeta(5)\right)\frac{1}{n}+\left(\frac{9}{2}\log(2)\zeta(3)-\frac{19}{240}\pi^4\right)\frac{1}{n^2}$$ $$\small+\left(\frac{1}{2}\log(2)\pi^2-6\zeta(3)\right)\frac{1}{n^3}+\left(3\log^2(2)-\frac{\pi^2}{2}\right)\frac{1}{n^4}+3\frac{ H_n^{(2)}}{n^4}+6\frac{H_n^{(3)}}{n^3}+9\frac{H_n^{(4)}}{n^2}+12\frac{H_n^{(5)}}{n}$$ $$\small-\frac{7}{120}\pi^4\frac{\overline{H}_n}{n}-\frac{9}{2}\zeta(3)\frac{\overline{H}_n}{n^2}-\frac{\pi^2}{2}\frac{\overline{H}_n}{n^3}-6\log(2)\frac{\overline{H}_n}{n^4}+3\frac{\overline{H}_n^2}{n^4}+6\frac{\overline{H}_n \overline{H}_n^{(2)}}{n^3}+6 \frac{\overline{H}_n \overline{H}_n^{(3)}}{n^2}+6\frac{\overline{H}_n \overline{H}_n^{(4)}}{n}$$ $$-\frac{9}{2}\zeta(3)\frac{\overline{H}_n^{(2)}}{n}-\frac{\pi^2}{2}\frac{\overline{H}_n^{(2)}}{n^2}-6\log(2)\frac{\overline{H}_n^{(2)}}{n^3}+3\frac{(\overline{H}_n^{(2)})^2}{n^2}+6\frac{\overline{H}_n^{(2)}\overline{H}_n^{(3)}}{n}-\frac{\pi^2}{2}\frac{\overline{H}_n^{(3)}}{n}$$ $$-6\log(2) \frac{\overline{H}_n^{(3)}}{n^2}-6\log(2) \frac{\overline{H}_n^{(4)}}{n}.$$

More details may be found in the paper Some Versatile Integrals with Parameter (2023)

Back to the main story

In a similar manner, we may establish similar relations with numbers of the type $\displaystyle \overline{O}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1}$ (maybe we might also find a name for these numbers). The extraction procedure is very similar to one of the skew-harmonic numbers previously shown, already described in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), Sect. 6.18. pages 592-595.

We start with the fact that $$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k+x}=\log(2)+2\frac{1}{x}-\psi(1+x)+\psi\left(\frac{x}{2}\right).$$

Proof: We have from (Almost) Impossible Integrals, Sums, and Series, Sect. 1.5, page 3, that $\displaystyle \int_0^1 \frac{t^{x-1}}{1+t}\textrm{d}t =\psi(x)-\psi\left(\frac{x}{2}\right)-\log(2)$, and if we combine it with the recurrence relation of Digamma function, $\displaystyle \psi(1+x)=\psi(x)+\frac{1}{x}$, we immediately obtain that
\begin{equation*} \psi(1+x)-\psi\left(\frac{x}{2}\right)-\frac{1}{x}-\log(2)= \int_0^1 \frac{t^{x-1}}{1+t}\textrm{d}t=\int_0^1 t^{x-1}\sum_{k=1}^{\infty}(-1)^{k-1} t^{k-1} \textrm{d}t \end{equation*} \begin{equation*} =\sum_{k=1}^{\infty}(-1)^{k-1}\int_0^1 t^{k+x-2} \textrm{d}t=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{1}{k+x-1}=\sum_{k=0}^{\infty}(-1)^k\frac{1}{k+x}=\frac{1}{x}-\sum_{k=1}^{\infty}(-1)^{k-1}\frac{1}{k+x}, \end{equation*} which proves the desires result.

Returning to the main result where we replace $x$ by $(2n-1)/2$, and then multiply both sides by $1/2$, we get that $$\frac{1}{2}\log(2)+2\frac{1}{2n-1}-\frac{1}{2}\psi\left(n+\frac{1}{2}\right)+\frac{1}{2}\psi\left(\frac{n}{2}-\frac{1}{4}\right)$$ $$=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{2(k+n)-1}=(-1)^n \sum_{k=n+1}^{\infty} (-1)^{k-1}\frac{1}{2k-1}=(-1)^n \underbrace{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{2k-1}}_{\displaystyle \pi/4}$$ $$+(-1)^{n-1} \sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1},$$ whence we obtain that $$\overline{O}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1}$$ $$=\frac{\pi}{4}+\frac{1}{2}\log(2)(-1)^{n-1} +2(-1)^{n-1} \frac{1}{2n-1}-\frac{1}{2}(-1)^{n-1} \psi\left(n+\frac{1}{2}\right)+\frac{1}{2}(-1)^{n-1} \psi\left(\frac{n}{2}-\frac{1}{4}\right),$$ where if we use the well-known fact that $\displaystyle \psi\left(n+\frac{1}{2}\right)=2 H_{2n}-H_n-\gamma-2\log(2)$, which, for example, is immediately extracted by using the identity $\displaystyle \psi(2 x)=\frac{1}{2}\psi(x)+\frac{1}{2}\psi\left(x+\frac{1}{2}\right)+\log(2)$, found and (very elegantly) derived in (Almost) Impossible Integrals, Sums, and Series (2019), pages 68-69, we arrive at

$$\overline{O}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1}$$ $$=\frac{\pi}{4}+\frac{1}{2}(\gamma+3\log(2))(-1)^{n-1} +2(-1)^{n-1} \frac{1}{2n-1}-(-1)^{n-1}H_{2n}$$ $$+\frac{1}{2}(-1)^{n-1}H_n+\frac{1}{2}(-1)^{n-1} \psi\left(\frac{n}{2}-\frac{1}{4}\right).$$

At this point, we can provide the desired result

$$\sum_{k=0}^n (-1)^{k-1}\frac{1}{2k+1}=-\overline{O}_n+(-1)^{n-1}\frac{1}{2n+1}$$ $$=-\frac{\pi}{4}-\frac{1}{2}(\gamma+3\log(2))(-1)^{n-1} -2(-1)^{n-1} \frac{1}{2n-1}+(-1)^{n-1} \frac{1}{2n+1}$$ $$+(-1)^{n-1}H_{2n}-\frac{1}{2}(-1)^{n-1}H_n-\frac{1}{2}(-1)^{n-1} \psi\left(\frac{n}{2}-\frac{1}{4}\right).$$

Final thoughts

We can also attack similarly the more general case $\displaystyle \overline{O}^{(m)}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{(2k-1)^m}$, and establish relations with Polygamma, but this will be part of another paper.

Having finished, for that Cauchy product, I would keep the notation involving $\overline{O}_n$ defined by $\displaystyle \overline{O}_n=\sum_{k=1}^n (-1)^{k-1}\frac{1}{2k-1}$, that is,

$$\frac{\operatorname{arctanh}(x)}{1+x^2}=\sum_{n=1}^{\infty} (-1)^{n-1}\overline{O}_n x^{2n-1},$$ which, in my opinion, is the most natural way to act.

End of story

Answer 3

$$\sum_{k=0}^n\frac{(-1)^k}{2k+1}=\sum_{k=0}^n(-1)^k\int_0^1 x^{2k}dx$$

$$=\int_0^1\sum_{k=0}^n(-x^2)^ndx$$

$$=\int_0^1\frac{1+(-1)^nx^{2n+2}}{1+x^2}dx$$

$$=\int_0^1\frac{1}{1+x^2}dx+(-1)^n\int_0^1\frac{x^{2n+2}}{1+x^2}dx$$

$$\overset{x=\sqrt{y}}{=}\frac{\pi}{4}+\frac{(-1)^n}{2}\int_0^1\frac{y^{n+\frac12}}{1+y}dy$$

$$=\frac{(-1)^n}{4}\left(H_{\frac{2n+1}{4}}-H_{\frac{2n-1}{4}}+(-1)^n\pi\right)$$

where the last result follows form using

$$ \int_0^1\frac{x^{s}}{1+x}dx=\frac12\left(H_{\frac s2}-H_{\frac{s-1}{2}}\right)$$

Thus,

$$\frac{\operatorname{arctanh}(x)}{1+x^2}=\sum_{n=0}^\infty (-1)^n\left(\sum_{k=0}^n \frac{(-1)^{k}}{2k+1}\right)x^{2n+1}$$

$$=\frac14\sum_{n=0}^\infty \left(H_{\frac{2n+1}{4}}-H_{\frac{2n-1}{4}}+(-1)^n\pi\right)x^{2n+1}$$ Shifting the index by $-1$, we get

$$\color{blue}{\frac{\operatorname{arctanh}(x)}{1+x^2}=\frac14\sum_{n=1}^\infty \left(H_{\frac{2n-1}{4}}-H_{\frac{2n-3}{4}}-(-1)^n\pi\right)x^{2n-1}}$$

By the way, replacing $x$ with $ix$ then using $\operatorname{arctanh}(ix)=i\arctan(x)$ gives

$$\color{green}{\frac{\arctan(x)}{1-x^2}=\frac14\sum_{n=1}^\infty (-1)^{n-1}\left(H_{\frac{2n-1}{4}}-H_{\frac{2n-3}{4}}-(-1)^n\pi\right)x^{2n-1}}$$