Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

One of the possible ways of computing the series is to obtain the generating function, but
this might be a tedious, hard work, pretty hard to obtain. What would you propose then?

$$\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$$

Solution 1:

Here is a solution that does not rely (too much) on softwares. I will be using the known values of the sums $\small{\displaystyle \sum^\infty_{n=1}\frac{H_n}{n2^n},\ \sum^\infty_{n=1}\frac{H_n}{n^22^n},\ \sum^\infty_{n=1}\frac{H_n}{n^32^n}}$.

Let $$\mathcal{S}=\sum^\infty_{n=1}\frac{H_n}{n^42^n}$$ We first consider a slightly different yet related sum. The main idea is to solve this sum with two different methods, one of which involves the sum in question. This then allows us to determine the value of the desired sum. \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4} =&\frac{1}{6}\sum^\infty_{n=1}(-1)^{n-1}H_n\int^1_0x^{n-1}\ln^3{x}\ {\rm d}x\\ =&\frac{1}{6}\int^1_0\frac{\ln^3{x}\ln(1+x)}{x(1+x)}{\rm d}x\\ =&\frac{1}{6}\int^1_0\frac{\ln^3{x}\ln(1+x)}{x}{\rm d}x-\frac{1}{6}\int^1_0\frac{\ln^3{x}\ln(1+x)}{1+x}{\rm d}x\\ =&\frac{1}{6}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\int^1_0x^{n-1}\ln^3{x}\ {\rm d}x-\frac{1}{6}\int^2_1\frac{\ln{x}\ln^3(x-1)}{x}{\rm d}x\\ =&\sum^\infty_{n=1}\frac{(-1)^{n}}{n^5}+\int^1_{\frac{1}{2}}\frac{\ln{x}\ln^3(1-x)}{6x}-\int^1_{\frac{1}{2}}\frac{\ln^2{x}\ln^2(1-x)}{2x}{\rm d}x\\&+\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln(1-x)}{2x}{\rm d}x-\int^1_{\frac{1}{2}}\frac{\ln^4{x}}{6x}{\rm d}x\\ =&-\frac{15}{16}\zeta(5)+\mathcal{I}_1-\mathcal{I}_2+\mathcal{I}_3-\mathcal{I}_4 \end{align} Starting with the easiest integral, \begin{align} \mathcal{I}_4=\frac{1}{30}\ln^5{2} \end{align} For $\mathcal{I}_3$, \begin{align} \mathcal{I}_3 =&-\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^1_{\frac{1}{2}}x^{n-1}\ln^3{x}\ {\rm d}x\\ =&-\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\frac{\partial^3}{\partial n^3}\left(\frac{1}{n}-\frac{1}{n2^n}\right)\\ =&\sum^\infty_{n=1}\left(\frac{3}{n^5}-\frac{3}{n^52^n}-\frac{3\ln{2}}{n^42^n}-\frac{3\ln^2{2}}{n^32^{n+1}}-\frac{\ln^3{2}}{n^22^{n+1}}\right)\\ =&3\zeta(5)-3{\rm Li}_5\left(\tfrac{1}{2}\right)-3{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}-\frac{3}{2}\ln^2{2}\left(\frac{7}{8}\zeta(3)-\frac{\pi^2}{12}\ln{2}+\frac{1}{6}\ln^3{2}\right)\\&-\frac{1}{2}\ln^3{2}\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\right)\\ =&3\zeta(5)-3{\rm Li}_5\left(\tfrac{1}{2}\right)-3{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}-\frac{21}{16}\zeta(3)\ln^2{2}+\frac{\pi^2}{12}\ln^3{2} \end{align} For $\mathcal{I}_2$, \begin{align} \mathcal{I}_2 =&\frac{1}{6}\ln^5{2}+\frac{1}{3}\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\ =&\frac{1}{6}\ln^5{2}-\frac{1}{3}\sum^\infty_{n=1}H_n\frac{\partial^3}{\partial n^3}\left(\frac{1}{n+1}-\frac{1}{(n+1)2^{n+1}}\right)\\ =&\frac{1}{6}\ln^5{2}+\sum^\infty_{n=1}\frac{2H_n}{(n+1)^4}-\sum^\infty_{n=1}\frac{2H_n}{(n+1)^42^{n+1}}-\sum^\infty_{n=1}\frac{2\ln{2}H_n}{(n+1)^32^{n+1}}\\ &-\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{(n+1)^22^{n+1}}-\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{3(n+1)2^{n+1}}\\ =&\frac{1}{6}\ln^5{2}+4\zeta(5)-\frac{\pi^2}{3}\zeta(3)-2\mathcal{S}+2{\rm Li}_5\left(\tfrac{1}{2}\right)-\frac{\pi^4}{360}\ln{2}+\frac{1}{4}\zeta(3)\ln^2{2}-\frac{1}{12}\ln^5{2}\\ &-\frac{1}{8}\zeta(3)\ln^2{2}+\frac{1}{6}\ln^5{2}-\frac{1}{6}\ln^5{2}\\ =&-2\mathcal{S}+2{\rm Li}_5\left(\tfrac{1}{2}\right)+4\zeta(5)-\frac{\pi^4}{360}\ln{2}+\frac{1}{8}\zeta(3)\ln^2{2}-\frac{\pi^2}{3}\zeta(3)+\frac{1}{12}\ln^5{2} \end{align} For $\mathcal{I}_1$, \begin{align} \mathcal{I}_1 =&\frac{1}{6}\int^{\frac{1}{2}}_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\ =&-\frac{1}{6}\sum^\infty_{n=1}H_n\frac{\partial^3}{\partial n^3}\left(\frac{1}{(n+1)2^{n+1}}\right)\\ =&\sum^\infty_{n=1}\frac{H_n}{(n+1)^42^{n+1}}+\sum^\infty_{n=1}\frac{\ln{2}H_n}{(n+1)^32^{n+1}}+\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{2(n+1)^22^{n+1}}+\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{6(n+1)2^{n+1}}\\ =&\mathcal{S}-{\rm Li}_5\left(\tfrac{1}{2}\right)+\frac{\pi^4}{720}\ln{2}-\frac{1}{16}\zeta(3)\ln^2{2}+\frac{1}{24}\ln^5{2} \end{align} Combining these four integrals as $\mathcal{I}_1-\mathcal{I}_2+\mathcal{I}_3-\mathcal{I}_4$ and $\displaystyle -\tfrac{15}{16}\zeta(5)$ gives \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4} =&3\mathcal{S}-6{\rm Li}_5\left(\tfrac{1}{2}\right)-\frac{31}{16}\zeta(5)-3{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}+\frac{\pi^4}{240}\ln{2}\\&-\frac{3}{2}\zeta(3)\ln^2{2}+\frac{\pi^2}{3}\zeta(3)+\frac{\pi^2}{12}\ln^3{2}-\frac{3}{40}\ln^5{2} \end{align} But consider $\displaystyle f(z)=\frac{\pi\csc(\pi z)(\gamma+\psi_0(-z))}{z^4}$. At the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{(-1)^n}{z^4(z-n)^2}+\frac{(-1)^nH_n}{z^4(z-n)}\right]\\ &=\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}+\frac{15}{4}\zeta(5) \end{align} At $z=0$, \begin{align} {\rm Res}(f,0) &=[z^3]\left(\frac{1}{z}+\frac{\pi^2}{6}z+\frac{7\pi^4}{360}z^3\right)\left(\frac{1}{z}-\frac{\pi^2}{6}z-\zeta(3)z^2-\frac{\pi^4}{90}z^3-\zeta(5)z^4\right)\\ &=-\zeta(5)-\frac{\pi^2}{6}\zeta(3) \end{align} At the negative integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,-n) &=\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}+\frac{15}{16}\zeta(5) \end{align} Since the sum of the residues is zero, $$\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}=-\frac{59}{32}\zeta(5)+\frac{\pi^2}{12}\zeta(3)$$ Hence, \begin{align} -\frac{59}{32}\zeta(5)+\frac{\pi^2}{12}\zeta(3) =&3\mathcal{S}-6{\rm Li}_5\left(\tfrac{1}{2}\right)-\frac{31}{16}\zeta(5)-3{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}+\frac{\pi^4}{240}\ln{2}\\&-\frac{3}{2}\zeta(3)\ln^2{2}+\frac{\pi^2}{3}\zeta(3)+\frac{\pi^2}{12}\ln^3{2}-\frac{3}{40}\ln^5{2} \end{align} This implies that \begin{align} \color{#FF4F00}{\sum^\infty_{n=1}\frac{H_n}{n^42^n}} \color{#FF4F00}{=}&\color{#FF4F00}{2{\rm Li}_5\left(\tfrac{1}{2}\right)+\frac{1}{32}\zeta(5)+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}-\frac{\pi^4}{720}\ln{2}+\frac{1}{2}\zeta(3)\ln^2{2}}\\&\color{#FF4F00}{-\frac{\pi^2}{12}\zeta(3)-\frac{\pi^2}{36}\ln^3{2}+\frac{1}{40}\ln^5{2}} \end{align} I will gladly provide a detailed solution for $\sum^\infty_{n=1}\frac{H_n}{n^32^n}$ too if there is a need.

Solution 2:

The sum is (with proof, see below) equal to $$ \def\tfrac#1#2{{\textstyle\frac{#1}{#2}}} 2 \text{Li}_5(\tfrac{1}{2})+\text{Li}_4(\tfrac{1}{2}) \log2-\tfrac{1}{2} \zeta (3) \zeta(2)+\tfrac{1}{32} \zeta (5)+\tfrac{1}{2} \zeta (3) \log^22-\tfrac{1}{6} \zeta (2) \log^32-\tfrac{1}{8} \zeta (4) \log(2)+\tfrac{1}{40} \log^52 $$

The sum is equal to $$ \def\Li{\mathrm{Li}} \Li_5(\tfrac12) + \zeta(-1,1,-1,1,1), $$ where $\zeta(-1,1,-1,1,1)$ is obtained by applying the multiple zeta function duality formula to the multiple polylogarithm sum $$ \sum_{i,j\geq1} \frac{2^{-i-j}}{i(i+j)^4} = \sum_{n\geq1}\frac{H_{n-1}}{2^nn^4} = \lambda\left({{4,1}\atop{2,2}}\right). $$ I think it is useful to write it in terms of a multiple polylogarithm sum, so that all the standard identities (Borwein, Bradley, Broadhurst, Lisonek, which I'll refer to as BBBL below) can be applied.

Another (I say very fitting) form for the sum is $$ 5\Li_5(\tfrac12)+\Li_4(\tfrac12)\log2-\frac16\int_1^\infty \frac{\log^3x\log(2x-1)}{x(2x-1)}\,dx, $$ where the integral is the integral representation (4.2 of BBBL) of $\lambda({4,1\atop2,2})$, integrated over one of the dimensions.

EDIT Okay, I found the identities now, so this is a proof. I will reference the BBBL paper I linked to above. The integral is, after $x\mapsto \frac12(1+1/t)$, $$ -\int_0^1 \frac{\log t}{t+1}\log^3\frac{t+1}{2t}, $$ which, after expanding the cube, doing some of the integrals with Mathematica, and expanding others in polylogarithms, as described here, becomes $$ 18\zeta(-4,1) + 6\zeta(-2,1,1,1) + 3\log^22\zeta(-2,1)-12\log2 \zeta(-3,1)+6\log2\zeta(-2,1,1) + 24\Li_5(\tfrac12) + 24\Li_4(\tfrac12)\log2 + \tfrac{81}{8}\zeta(5)-6\zeta(2)\zeta(3)+15\zeta(3)\log^22+\tfrac45\log^52+\tfrac45\log^52-\tfrac34\pi^2\log^32-\tfrac7{40}\pi^4\log2. $$ The "easy" integrals here were done by Mathematica. The closed forms for $\zeta(-s,1) = \alpha_h(1,s)$ Mathematica doesn't know. The other unknown terms are $\zeta(-2,1,1,1)$ and $\zeta(-2,1,1)$. Using Theorem 9.3 of BBBL, and then Theorem 8.3 and Corollary 1, these are $$\begin{eqnarray} \zeta(-2,1,1,1) &=& \mu(\{-1\}^4,1) - \mu(\{-1\}^5) \\&=& -\text{Li}_5(\tfrac{1}{2})-\text{Li}_4(\tfrac{1}{2}) \log2+\zeta (5)-\tfrac{7}{16} \zeta (3) \log^22+\tfrac{1}{6}\zeta (2) \log^32+\tfrac{1}{30} (-\log^52) \\ \zeta(-2,1,1) &=& \mu(\{-1\}^3,1) - \mu(\{-1\}^4) \\&=& \text{Li}_4(\tfrac{1}{2})+\tfrac{7}{8} \zeta (3) \log2-\zeta (4)-\tfrac{1}{4} \zeta (2) \log^22+\tfrac{1}{24} \log^42 \end{eqnarray}$$

Each sum $\zeta(-s,1)=\sum_{k\geq1}H_{k-1}(-1)^k/k^s$ is already known, for even $s$, or odd $s\leq3$, see Flajolet and Salvy: $$\begin{eqnarray} \zeta(-2,1) &=& \tfrac18\zeta(3) \\ \zeta(-3,1) &=& 2 \text{Li}_4(\tfrac{1}{2})+\tfrac{7}{4} \zeta (3) \log(2)-\tfrac{15}{8} \zeta (4)-\tfrac{1}{2} \zeta (2) \log^2(2)+\tfrac{1}{12} \log^42 \\ \zeta(-4,1) &=& \tfrac{1}{2} \zeta (3) \zeta (2)-\tfrac{29}{32} \zeta (5) \end{eqnarray}$$

So, the integral equals $$ 18 \text{Li}_5(\tfrac{1}{2})+3 \zeta (3) \zeta (2)-\tfrac{3}{16} \zeta (5)-3 \zeta (3) \log^22+\zeta (2) \log^3(2)+\tfrac{3}{4} \zeta (4) \log2+\tfrac{3}{20} (-\log^52) $$

Putting together gives the form I got numerically as well.

Solution 3:

Different approach using only real analysis to prove the following equality:

\begin{align} \displaystyle\sum_{n=1}^{\infty}\frac{H_n}{2^n n^4}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2) +\frac12\ln^22\zeta(3)\\ &\quad-\frac18\ln2\zeta(4)- \frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52 \end{align}

Proof: Using the algebraic identity: $$ 6a^2b^2-4ab^3=(a-b)^4+4a^3b-b^4-a^4 $$ and letting $a=\ln x$, $b=\ln(1-x)$ we get \begin{equation*} 6\ln^2x\ln^2(1-x)-4\ln x\ln^3(1-x)=\ln^4\left( \frac{x}{1-x}\right) +4\ln^3x\ln(1-x)-\ln^4(1-x)-\ln^4x \end{equation*} Dividing both sides by $ x $ then integrating from $ x=1/2 $ to $ 1 $ we have: \begin{align*} I&=6\int_{1/2}^{1}\frac{\ln^2x\ln^2(1-x)}{x}\,dx-4\int_{1/2}^{1}\frac{\ln x\ln^3(1-x)}{x}\,dx\\ &=\int_{1/2}^{1}\frac{1}{x}\ln^4\left(\frac{x}{1-x}\right)\ dx+4\int_{1/2}^{1}\frac{\ln^3x\ln(1-x)}{x}\,dx-\int_{1/2}^{1}\frac{\ln^4(1-x)}{x}\ dx-\int_{1/2}^{1}\frac{\ln^4x}{x}\ dx\\ I&=6I_1-4I_2=I_3+4I_4-I_5-\frac15\ln^52 \end{align*}

The first and second integrals: Applying IBP for the first integral by setting $ dv=\frac{\ln^2x}{x} $ and $ u=\ln^2(1-x) $ and letting $ x\mapsto 1-x $ for the second integral, we get: \begin{align*} I&=2\ln^52+4\int_{1/2}^{1}\frac{\ln^3x\ln(1-x)}{1-x}\,dx-4\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{1-x}\,dx\\ \tag{$ i $} &=2\ln^52+4\int_{0}^{1}\frac{\ln^3x\ln(1-x)}{1-x}\,dx-8\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{1-x}\,dx\\ \tag{$ ii $} &=\small{2\ln^52-4\sum_{n=1}^{\infty}\left( H_n-\frac{1}{n}\right) \int_0^1 x^{n-1}\ln^3x\,dx+8\sum_{n=1}^{\infty}\left( H_n-\frac{1}{n}\right) \int_0^{1/2} x^{n-1}\ln^3x\,dx}\\ &=\small{2\ln^52-24\zeta(5)+24\sum_{n=1}^{\infty}\frac{H_n}{n^4}+8\sum_{n=1}^{\infty}H_n\int_{0}^{1/2}x^{n-1}\ln^3x\ dx-8\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1/2}x^{n-1}\ln^3x\ dx}\tag{1} \end{align*} note that in $ (i) $ we used $ \int_{1/2}^{1}f(x)\,dx = \int_{0}^{1}f(x)\,dx- \int_{0}^{\tiny{1/2}}f(x)\,dx$ and in $ (ii) $ we used $ \frac{\ln(1-x)}{1-x}=-\sum_{n=1}^{\infty}H_n x^n=-\sum_{n=1}^{\infty}\left(H_n-\frac{1}{n}\right) x^{n-1} $

The third integral: Using the change of variable $ x=\frac{1}{1+y} $ we get \begin{align*} I_3&=\int_{1/2}^{1}\frac1x\ln^4\left( \frac{x}{1-x}\right)\ dx=\int_0^1\frac{\ln^4x}{1+x}\,dx=-\sum_{n=1}^{\infty}(-1)^n\int_0^1 x^{n-1}\ln^4x\,dx\\ &=-24\sum_{n=1}^{\infty}\frac{(-1)^n}{n^5}=-24\operatorname{Li_5}(-1)=\frac{45}{2}\zeta(5) \end{align*} The forth integral: \begin{align*} I_4&=\int_{1/2}^{1}\frac{\ln^3x\ln(1-x)}{x}\,dx=\int_{0}^{1}\frac{\ln^3x\ln(1-x)}{x}\,dx-\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{x}\,dx\\ &=-\sum_{n=1}^{\infty}\frac1n \int_0^1 x^{n-1}\ln^3x\,dx-\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{x}\,dx =6\zeta(5)-\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{x}\,dx \end{align*}

The fifth integral: Applying IBP by setting $ dv=\frac1x $ and $ u=\ln^4(1-x)$ we have \begin{align} I_5&=\int_{1/2}^{1}\frac{\ln^4(1-x)}{x}\,dx=\ln^52+4\underbrace{\int_{1/2}^{1}\frac{\ln x\ln^3(1-x)}{1-x}\,dx}_{\displaystyle\small{x\mapsto 1-x}}\\ &=\ln^52+4\int_{0}^{1/2}\frac{\ln(1-x)\ln^3x}{x}\,dx \end{align}

Grouping $ I_3,I_4 $ and $ I_5 $ we have \begin{align*} I&=\frac{93}{2}\zeta(5)-\frac65\ln^52-8\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{x}\,dx\\ &=\frac{93}{2}\zeta(5)-\frac65\ln^52+8\sum_{n=1}^{\infty}\frac1n\int_{0}^{1/2}x^{n-1}\ln^3x\,dx \tag{2} \end{align*} Combining $ (1) $ and $ (2) $ we have \begin{equation*} \sum_{n=1}^{\infty}H_n\int_{0}^{1/2}x^{n-1}\ln^3x\,dx=\frac{141}{16}\zeta(5)-\frac25\ln^52-3\sum_{n=1}^{\infty}\frac{H_n}{n^4}+2\sum_{n=1}^{\infty}\frac1n\int_{0}^{1/2}x^{n-1}\ln^3x\,dx \end{equation*} since

$$-\int_{0}^{1/2}x^{n-1}\ln^3x\,dx= \frac{\ln^32}{2^n n}+\frac{3\ln^22}{2^n n^2}+\frac{6\ln2}{2^n n^3}+\frac{6}{2^n n^4}$$

then

$$-\sum_{n=1}^{\infty}H_n\left( \frac{\ln^32}{2^n n}+\frac{3\ln^22}{2^n n^2}+\frac{6\ln2}{2^n n^3}+\frac{6}{2^n n^4}\right)\\=\frac{141}{16}\zeta(5)-\frac25\ln^52-3\sum_{n=1}^{\infty}\frac{H_n}{n^4}-2\sum_{n=1}^{\infty}\frac1n\left( \frac{\ln^32}{2^n n}+\frac{3\ln^22}{2^n n^2}+\frac{6\ln2}{2^n n^3}+\frac{6}{2^n n^4}\right)$$

Thus, rearranging the terms and simplifying we have \begin{align*} \sum_{n=1}^{\infty}\frac{H_n}{2^nn^4} &=-\ln2\sum_{n=1}^{\infty}\frac{H_n}{2^n n^3}-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{2^n n^2}-\frac16\ln^32\sum_{n=1}^{\infty}\frac{H_n}{2^n n}+\frac12\sum_{n=1}^{\infty}\frac{H_n}{n^4}-\frac{47}{32}\zeta(5)\\ &\quad+\frac{1}{15}\ln^52+\frac{1}{3}\ln^32\operatorname{Li_2}\left( \frac12\right)+\ln^22\operatorname{Li_3}\left( \frac12\right)+2\ln2\operatorname{Li_4}\left( \frac12\right) +2\operatorname{Li_5}\left( \frac12\right) \end{align*} Substituting the value of the first sum and the second sum gives our desired closed form.

note that $ \operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$ and $ \operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$

Solution 4:

The following new solution to the classical result, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}=\frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3)$, is proposed by Cornel Ioan Valean, using a very simple real technique based upon the powerful identity, $$\sum _{k=1}^{\infty } \frac{1}{2k(2k+2n-1)}=\frac{1}{2(2n-1)}\left(2H_{2n}-H_n-2\log(2)\right),\tag1$$ found and proved in $(6.289)$ in the book (Almost) Impossible Integrals, Sums, and Series. The solution may also be easily extended to calculate the generalization, $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n^{2m}}$.

Upon multiplying both sides of $(1)$ by $1/(2n-1)^3$, summing from $n=1$ to $\infty$ and then reindexing, we have for the right-hand side that $$\sum_{n=1}^{\infty} \frac{H_{2n}}{(2n-1)^4}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{H_n}{(2n-1)^4}-\log(2)\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}$$ $$=-\frac{15}{16}\log(2)\zeta(4)+\sum_{n=1}^{\infty} \frac{H_{2n-1}}{(2n-1)^4}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{H_n}{(2n+1)^4}$$ $$=\frac{21 }{32}\zeta (2) \zeta (3)-\frac{31 }{16}\zeta (5)+\frac{1}{2}\sum _{n=1}^{\infty } \frac{H_n}{n^4}+\frac{1}{2}\sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_n}{n^4}$$ $$=\frac{5}{32}\zeta(2)\zeta(3)-\frac{7}{16}\zeta(5)+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_{n}}{n^4}.\tag2$$

On the other hand, based on $(1)$, we have for the left-hand side that $$\sum _{n=1}^{\infty}\left(\sum _{k=1}^{\infty } \frac{1}{2k(2k+2n-1)(2n-1)^3}\right)=\sum _{k=1}^{\infty}\left(\sum _{n=1}^{\infty } \frac{1}{2k(2k+2n-1)(2n-1)^3}\right)$$ $$=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2} \sum _{n=1}^{\infty } \frac{1}{(2n-1)^3}-\frac{1}{8}\sum _{k=1}^{\infty}\frac{1}{k^3} \sum _{n=1}^{\infty } \frac{1}{(2n-1)^2}+\frac{1}{16}\sum _{k=1}^{\infty}\frac{1}{k^4}\sum _{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{2n+2k-1}\right)$$ $$=\frac{1}{8}\zeta(2)\zeta(3)+\frac{1}{16}\sum_{k=1}^{\infty}\frac{1}{k^4}\sum_{n=1}^k\frac{1}{2n-1}=\frac{1}{8}\zeta(2)\zeta(3)+\frac{1}{16}\sum_{k=1}^{\infty}\frac{1}{k^4}\left(H_{2k}-\frac{1}{2}H_k\right)$$ $$=\frac{1}{8}\zeta(2)\zeta(3)-\frac{1}{32}\sum_{k=1}^{\infty}\frac{H_k}{k^4}+\sum_{k=1}^{\infty}\frac{H_{2k}}{(2k)^4}=\frac{5}{32}\zeta(2)\zeta(3)-\frac{3}{32}\zeta(5)+\sum_{k=1}^{\infty}\frac{H_{2k}}{(2k)^4}$$ $$=\frac{5}{32}\zeta(2)\zeta(3)-\frac{3}{32}\zeta(5)+\frac{1}{2}\sum_{k=1}^{\infty}\frac{H_{k}}{k^4}-\frac{1}{2}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{H_{k}}{k^4}$$ $$=\frac{45}{32}\zeta(5)-\frac{11}{32}\zeta(2)\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}.\tag3$$

Combining $(2)$ and $(3)$, we obtain $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}=\frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3).$$

In the calculations we needed particular cases of the generalizations, \begin{equation*} 2\sum_{k=1}^\infty \frac{H_k}{k^n}=(n+2)\zeta(n+1)-\sum_{k=1}^{n-2} \zeta(n-k) \zeta(k+1), \ n\ge2, \end{equation*} and \begin{equation*} \sum _{k=1}^{\infty}\frac{H_k}{(2k+1)^{2m}}=2m\left(1-\frac{1}{2^{2m+1}}\right)\zeta(2m+1)-2\log(2)\left(1-\frac{1}{2^{2m}}\right)\zeta(2m) \end{equation*} \begin{equation*} -\frac{1}{2^{2m}}\sum_{i=1}^{m-1}(1-2^{i+1})(1-2^{2m-i})\zeta(1+i)\zeta(2m-i), \end{equation*} proved in https://math.stackexchange.com/q/3268851. Combining the chosen answer with this one we obtain another evaluation by real methods of the series $\displaystyle \sum_{n=1}^{\infty}\frac{H_{ n}}{2^nn^4}$.

Cornel has also prepared an article with the generalization $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n^{2m}}$ which is available here (note these series are usually very hard to evaluate by real methods exclusively).