Approaching $\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}$ elegantly

Question

How to elegantly prove that $$\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}=2\text{Li}_4\left(\frac12\right)-\frac{49}{16}\zeta(4)+\frac72\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number, $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number, $\text{Li}_r$ is the polylogarithm function and $\zeta$ is the Riemann zeta function.

What I mean by elegant solutions is solutions involving cancellation of challenging integrals/ sums , symmetry , manipulations and new ideas that save us tedious calculations. However, all solutions are appreciated.

Thank you

Answer 1

Here's a preliminary answer which boils the problem to find the sum

$$s = \sum_{n=1}^{\infty} \frac{1}{n^3}(\overline{H_{n}} - H_{n/2})\tag{1}$$

down to the tough (?) sum

$$s_1 = -\sum_{n=1}^{\infty} \frac{H_{n-\frac{1}{2}}}{(2n-1)^3}\tag{2}$$

Let us, just for information, look at the integral representation of the sum

$$s_i = \int_{0}^{1} \sum _{n=1}^{\infty } \frac{\frac{1-(-x)^n}{x+1}-\frac{1-x^{n/2}}{1-x}}{n^3}\,dx \\= \int_{0}^{1}\frac{-x \operatorname{Li}_3\left(\sqrt{x}\right)-\operatorname{Li}_3\left(\sqrt{x}\right)-x \text{Li}_3(-x)+\text{Li}_3(-x)+2 x \zeta (3)}{(x-1) (x+1)}\,dx \\\simeq 0.260631\tag{3}$$

The main idea is to split the sum $(1)$ into even and odd parts and then use the well-known relations

$$\overline{H_{2k}} = H_{2k} - H_{k}, \overline{H_{2k+1}}=H_{2k+1} - H_{k}\tag{4a}$$

and

$$\overline{H_{2k-1}}=H_{2k-1}-H_{k}+\frac{1}{n}\tag{4b}$$

This gives

$$s = s_1 + s_2 + s_3+ s_4 + s_5 + s_6$$

Where

$\begin{align} &s_2 = \sum_{n=1}^{\infty}\frac{H_{2n}}{(2n)^3}\\ &s_3 = \sum_{n=1}^{\infty}\frac{H_{2n-1}}{(2n-1)^3}\\ &s_4 = -2\sum_{n=1}^{\infty}\frac{H_{n}}{(2n)^3}\\ &s_5 =- \sum_{n=1}^{\infty}\frac{H_{n}}{(2n-1)^3}\\ &s_6 = \sum_{n=1}^{\infty}\frac{1}{n(2n-1)^3} \end{align}$

Notice that

$$s_2+s_3= \sum_{n=1}^{\infty}\frac{H_{n}}{n^3}$$

and

$$s_A = s_2+s_3+s_4+s_5 = \sum _{n=1}^{\infty } \left(\frac{1}{n^3}-\frac{1}{(2 n)^3}-\frac{1}{(2 n-1)^3}\right) H_n\tag{5}$$

so that

$$s = s_1 + s_A + s_6\tag{6}$$

Mathematica gives

$$s_A =-\frac{7 \pi ^4 \zeta (3)}{720}+40 \zeta (3)-\frac{7 \pi ^2 \zeta (5)}{48}+\frac{7 \zeta (7)}{2}+14 \zeta (3) \log (2) \\ +8 \pi ^2-\frac{\pi ^4}{9}+48 \log ^2(2)-6 \pi ^2 \log (2)-160 \log (2)\tag{7}$$

and

$$s_6 = \frac{7 \zeta (3)}{4}-\frac{\pi ^2}{4}+\log (4)\tag{8}$$

The result $(6)$ is numerically correct.

I am sure that someone around here has already calculated the sum $s_1$ which would then complete the result.

Answer 2

In this solution we have

$$\small{\sum_{n=1}^\infty\frac{H_{n/2}}{n}x^n-\sum_{n=1}^\infty \frac{\overline{H}_n}{n}x^n=2 \text{Li}_2(x)+\text{Li}_2(-x)+\frac{1}{2} \ln ^2(1-x^2)-\frac{1}{2} \ln ^2(1+x)+\ln(2)\ln\left(\frac{1-x}{1+x}\right)}$$

Multiply both sides by $\frac{\ln x}{x}$ then $\int_0^1$ and use the fact that $\int_0^1 x^{n-1}\ln x\ dx=-\frac{1}{n^2}$ we get

$$S=\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}=2\int_0^1\frac{\ln x\text{Li}_2(x)}{x}\ dx+\int_0^1\frac{\ln x\text{Li}_2(-x)}{x}\ dx$$

$$+\frac12\int_0^1\frac{\ln x\ln^2(1-x^2)}{x}-\frac12\int_0^1\frac{\ln x\ln^2(1+x)}{x}+\ln(2)\int_0^1\frac{\ln x\ln\left(\frac{1-x}{1+x}\right)}{x}\ dx$$

Lets calculate each integral

$$\int_0^1\frac{\ln x\text{Li}_2(x)}{x}\ dx=\sum_{n=1}^\infty \frac1{n^2}\int_0^1 x^{n-1}\ln x\ dx=-\sum_{n=1}^\infty \frac1{n^4}=-\zeta(4)$$

$$\int_0^1\frac{\ln x\text{Li}_2(-x)}{x}\ dx=\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln x\ dx=-\sum_{n=1}^\infty \frac{(-1)^n}{n^4}=\frac78\zeta(4)$$

$$\int_0^1\frac{\ln x\ln^2(1-x^2)}{x}\ dx=\frac14\int_0^1\frac{\ln x\ln^2(1-x)}{x}\ dx=\frac12\sum_{n=1}^\infty \frac{H_{n-1}}{n}\int_0^1 x^{n-1}\ln x\ dx$$

$$=-\frac12\sum_{n=1}^\infty \frac{H_{n-1}}{n^3}=-\frac12\sum_{n=1}^\infty \frac{H_{n}}{n^3}+\frac12\zeta(4)=-\frac1{8}\zeta(4)$$

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

$$=-2\sum_{n=1}^\infty \frac{(-1)^n H_{n-1}}{n^3}=-2\sum_{n=1}^\infty \frac{(-1)^n H_{n}}{n^3}-\frac74\zeta(4)$$

$$=\frac{15}{4}\zeta(4)-\frac72\ln(2)\zeta(3)+\ln^2(2)\zeta(2)-\frac16\ln^4(2)-4\text{Li}_4\left(\frac12\right)$$

$$\int_0^1\frac{\ln x\ln\left(\frac{1-x}{1+x}\right)}{x}\ dx\overset{IBP}{=}\int_0^1\frac{\ln^2x}{1-x^2}\ dx=\sum_{n=0}^\infty \int_0^1 x^{2n}\ln^2x\ dx$$ $$=\sum_{n=0}^\infty\frac{2}{(2n+1)^3}=\frac74\zeta(3)$$

Combine all these results, the closed form of $S$ follows.

Note that $\sum_{n=1}^\infty\frac{H_n}{n^3}$ can be obtained using Euler identity and $\sum_{n=1}^\infty\frac{(-1^n) H_n}{n^3}$ is calculated here.

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Bonus

I am going to evaluate $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}$ without using any generalization.

From above we have

$$\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}=2\underbrace{\int_0^1\frac{\ln x\text{Li}_2(x)}{x}\ dx}_{-\zeta(4)}+\underbrace{\int_0^1\frac{\ln x\text{Li}_2(-x)}{x}\ dx}_{7/8\zeta(4)}$$

$$+\frac12\underbrace{\int_0^1\frac{\ln x\ln^2(1-x^2)}{x}}_{-1/8\zeta(4)}-\frac12\underbrace{\int_0^1\frac{\ln x\ln^2(1+x)}{x}}_{-2\sum_{n=1}^\infty \frac{(-1)^n H_{n}}{n^3}-\frac74\zeta(4)}+\ln(2)\underbrace{\int_0^1\frac{\ln x\ln\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{7/4\zeta(3)}$$

Also its easy to prove that $\sum_{n=1}^\infty \frac{H_{n/2}}{n^3}=-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$.

Notice that $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$ magically cancels out from both sides and we are left with

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}=\frac74\ln2\zeta(3)-\frac5{16}\zeta(4)$$

Answer 3

Expanding on my comment above:

Let $\mathcal{S}$ denote the value the following infinite series:

$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\overline{H}_{n}-H_{n/2}}{n^{3}}\approx0.260631,$$

where $\overline{H}_{n}$ here denotes the $n$-th alternating harmonic number and is defined for each positive integer $n$ by the finite series

$$\overline{H}_{n}:=\sum_{k=1}^{n}\frac{\left(-1\right)^{k-1}}{k};~~~\small{n\in\mathbb{N}},$$

and the $\alpha$-th harmonic number $H_{\alpha}$ is defined here for real argument $\alpha$ through Euler's integral representation

$$H_{\alpha}:=\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{\alpha}}{1-t};~~~\small{\alpha\in\left(-1,\infty\right)}.$$

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An integral representation for the alternating harmonic numbers can be obtained as follows: for $n\in\mathbb{N}$, we have

$$\begin{align} \overline{H}_{n} &=\sum_{k=1}^{n}\frac{\left(-1\right)^{k-1}}{k}\\ &=\sum_{k=1}^{n}\left(-1\right)^{k-1}\int_{0}^{1}\mathrm{d}t\,t^{k-1}\\ &=\int_{0}^{1}\mathrm{d}t\,\sum_{k=1}^{n}\left(-t\right)^{k-1}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1-\left(-t\right)^{n}}{1+t}.\\ \end{align}$$

The difference $\overline{H}_{n}-H_{n/2}$ can be expressed as a single definite integral by combining the representations above:

$$\begin{align} \overline{H}_{n}-H_{n/2} &=\int_{0}^{1}\mathrm{d}x\,\frac{1-\left(-x\right)^{n}}{1+x}-\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{n/2}}{1-t}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1-\left(-x\right)^{n}}{1+x}-\int_{0}^{1}\mathrm{d}x\,\frac{2x\left(1-x^{n}\right)}{1-x^{2}};~~~\small{\left[t=x^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1-\left(-x\right)^{n}}{1+x}-\frac{2x\left(1-x^{n}\right)}{1-x^{2}}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1-\left(-x\right)^{n}}{1+x}-\frac{1-x^{n}}{1-x}+\frac{1-x^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{2}{1+x}-\frac{1-x^{n}}{1-x}-\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right].\\ \end{align}$$

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Recall that for positive integer order $p$ and complex argument $z$, the $p$-th order polylogarithm $\operatorname{Li}_{p}{\left(z\right)}$ is defined on the unit disk by the infinite series

$$\operatorname{Li}_{p}{\left(z\right)}:=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{p}};~~~\small{p>1\land\left|z\right|\le1}.$$

Using the technique of switching the order of summation and integration, it's then a straightforward matter to convert the series representation for $\mathcal{S}$ into a polylogarithmic integral. We find

$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}\frac{\overline{H}_{n}-H_{n/2}}{n^{3}}\\ &=\sum_{n=1}^{\infty}\frac{1}{n^{3}}\int_{0}^{1}\mathrm{d}x\,\left[\frac{2}{1+x}-\frac{1-x^{n}}{1-x}-\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\sum_{n=1}^{\infty}\frac{1}{n^{3}}\left[\frac{2}{1+x}-\frac{1-x^{n}}{1-x}-\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\frac{2}{1+x}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\frac{1-x^{n}}{1-x}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{2\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}-\frac{\operatorname{Li}_{3}{\left(x\right)}+\operatorname{Li}_{3}{\left(-x\right)}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}+\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-x\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\right].\\ \end{align}$$

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Define the function $F:\left(-\infty,1\right]\rightarrow\mathbb{R}$ via the polylogarithmic expression

$$F{\left(x\right)}:=\frac12\left[\operatorname{Li}_{2}{\left(x\right)}\right]^{2}-\ln{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right];~~~\small{x<1},$$

$$F{\left(1\right)}:=\lim_{x\to1^{-}}\bigg{[}\frac12\left[\operatorname{Li}_{2}{\left(x\right)}\right]^{2}-\ln{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right]\bigg{]}=\frac12\left[\operatorname{Li}_{2}{\left(1\right)}\right]^{2}.$$

Calculating the derivative of $F$, we find

$$\begin{align} \frac{d}{dx}F{\left(x\right)} &=\frac{d}{dx}\bigg{[}\frac12\left[\operatorname{Li}_{2}{\left(x\right)}\right]^{2}-\ln{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right]\bigg{]}\\ &=-\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}-\ln{\left(1-x\right)}\left[-\frac{\operatorname{Li}_{2}{\left(x\right)}}{x}\right]+\frac{1}{1-x}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right]\\ &=\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x},\\ \end{align}$$

i.e., $F{\left(x\right)}$ is an antiderivative of $\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}$, allowing us to reduce the integral form for $\mathcal{S}$ to

$$\begin{align} \mathcal{S} &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}+\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-x\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}+\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-x\right)}}{1+x}\right]-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\operatorname{Li}_{3}{\left(1\right)}\ln{\left(1+x\right)}-F{\left(x\right)}-F{\left(-x\right)}\right]-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\operatorname{Li}_{3}{\left(1\right)}\ln{\left(2\right)}-F{\left(1\right)}-F{\left(-1\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\operatorname{Li}_{3}{\left(1\right)}\ln{\left(2\right)}-\frac12\left[\operatorname{Li}_{2}{\left(1\right)}\right]^{2}-\frac12\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}+\ln{\left(2\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-1\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\frac{11}{4}\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac58\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}.\\ \end{align}$$

It remains to evaluate the integral $\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}$. Now, there is an extremely tidy way to calculate this integral in terms of Nielsen generalized polylogarithms:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x};~~~\small{I.B.P.s}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(-x\right)}}{x};~~~\small{I.B.P.s}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x}\int_{0}^{1}\mathrm{d}y\,\frac{(-1)\ln{\left(1+xy\right)}}{y}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-x\right)}\ln{\left(1+xy\right)}}{xy}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\ln{\left(1+xy\right)}}{xy}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\,\frac{1}{y}\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\ln{\left(1+yx\right)}}{x}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\,\frac{1}{y}\left[\operatorname{Li}_{3}{\left(-y\right)}+S_{1,2}{\left(-y\right)}\right]\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\,\frac{d}{dy}\left[\operatorname{Li}_{4}{\left(-y\right)}+S_{2,2}{\left(-y\right)}\right]\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\left[\operatorname{Li}_{4}{\left(-1\right)}+S_{2,2}{\left(-1\right)}\right]\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}+\frac78\zeta{\left(4\right)}-S_{2,2}{\left(-1\right)}.\\ \end{align}$$

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To summarize, the evaluation of the series $\mathcal{S}$ can be boiled down to finding a closed-form expression for the Nielsen generalized polylogarithmic term $S_{2,2}{\left(-1\right)}$.