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

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 $S_{2,2}{\left(-1\right)}$.

Answer 2

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 3

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 4

A solution in large steps from Cornel

Supposing it's not too late to join the party, the most powerful approach (in my opinion) is to exploit Digamma involving half the argument as presented in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), page $417-419$, Sect. $4.18$, which is the sequel of (Almost) Impossible Integrals, Sums, and Series (2019), that is, $$\displaystyle \psi\left(\frac{n}{2}\right)=-\log(2)-\gamma-\log(2)(-1)^{n-1}-2\frac{1}{n}+H_n+(-1)^{n-1} \overline{H}_n,$$ or $$\displaystyle \psi\left(\frac{n}{2}+1\right)=-\log(2)-\gamma-\log(2)(-1)^{n-1}+H_n+(-1)^{n-1} \overline{H}_n. \tag1$$

Since we also know that $H_n=\psi(n+1)+\gamma$, in view of, say, $(1)$, we have the transformation

$$\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}$$ $$=\sum_{n=1}^\infty\frac{(1-(-1)^{n-1})\overline{H}_n}{n^3}+\log(2)\sum_{n=1}^\infty\frac{1}{n^3}+\log(2)\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}-\sum_{n=1}^\infty\frac{H_n}{n^3}$$ $$\{\text{recall that } \overline{H}_{2n}= H_{2n}-H_n \}$$ $$=\frac{1}{4}\sum_{n=1}^\infty\frac{H_{2n}-H_n}{n^3}+\log(2)\sum_{n=1}^\infty\frac{1}{n^3}+\log(2)\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}-\sum_{n=1}^\infty\frac{H_n}{n^3},$$ and upon expanding, we notice at this point all series are known.

Alternatively, you can exploit the parity.

End of story {the power of these relations makes a world (or a universe) of difference}