TITLE: prove $\sum\limits_{n\geq 1} (-1)^{n+1}\frac{H_{\lfloor n/2\rfloor}}{n^3} = \zeta^2(2)/2-\frac{7}{4}\zeta(3)\log(2)$
QUESTION [16 upvotes]: Prove the following 
$$\sum\limits_{n\geq 1}(-1)^{n+1}\frac{H_{\lfloor n/2\rfloor}}{n^3} = \frac{1}{2}\zeta(2)^2-\frac{7}{4}\zeta(3)\log(2)$$
I was able to prove the formula above and interested in what approach you would take .

REPLY [10 votes]: Here is another variation of the theme: We start by splitting the sum
\begin{align*}
\sum_{n\geq 1}(-1)^{n+1}\frac{H_{\lfloor n/2\rfloor}}{n^3}&=-\sum_{n\geq1}\frac{H_n}{(2n)^3}+\sum_{n\geq1}\frac{H_n}{(2n+1)^3}\\
&=-\frac{1}{8}\sum_{n\geq1}\frac{H_n}{n^3}+\sum_{n\geq1}\frac{H_n}{(2n+1)^3}
\end{align*}
According to the answers of question 305476

the following identity is valid:
  \begin{align*}
\sum_{n\geq1}\frac{H_n}{n^3}=\frac{1}{2}\zeta(2)^2=\frac{\pi^4}{72}
\end{align*}

A very elegant answer was given by the user Random Variable. It was based on two ideas. The first one was using the integral representation
\begin{align*}
H_n=\int_0^1\frac{1-t^n}{1-t}dt
\end{align*}
for the harmonic numbers $H_n$ in order to simplify the sum. The second idea was using integration by parts and so switching from expressions with $\text{Li}_3$ to expressions with $\text{Li}_2$ reducing thereby the complexity of the integral. 
Now, we use the same approach with some additional adaptations to show that

the following identity is valid:
  \begin{align*}
\sum_{n\geq1}\frac{H_n}{(2n+1)^3}&=\frac{9}{16}\zeta(2)^2-\frac{7}{4}\zeta(3)\ln(2)\\
&=\frac{\pi^4}{64}-\frac{7}{4}\zeta(3)\ln(2)
\end{align*}

We start by using the integral representation for $H_n$ from above and the substitution $u^2=t$:

\begin{align*}
\sum_{n\geq1}\frac{H_n}{(2n+1)^3}&=\int_0^1\frac{1}{1-t}\sum_{n\geq1}\frac{1-t^n}{(2n+1)^3}dt\\
&=\int_0^1\frac{1}{1-u^2}\sum_{n\geq1}\frac{1-u^{2n}}{(2n+1)^3}2udu
\end{align*}

Now observe, that
\begin{align*}
\sum_{n\geq1}\frac{1}{(2n+1)^3}&=\sum_{n\geq1}\frac{1}{n^3}-\sum_{n\geq1}\frac{1}{(2n)^3}-1
=\frac{7}{8}\zeta(3)-1\\
\sum_{n\geq1}\frac{u^{2n+1}}{(2n+1)^3}&=\sum_{n\geq1}\frac{u^n}{n^3}-\sum_{n\geq1}\frac{u^{2n}}{(2n)^3}-u=\text{Li}_3(u)-\frac{1}{8}\text{Li}_3(u^2)-u\\
\end{align*}

Therefore we get
  \begin{align*}
\sum_{n\geq1}\frac{H_n}{(2n+1)^3}&=2\int_0^1\frac{1}{1-u^2}\left(\frac{7}{8}\zeta(3)u-\text{Li}_3(u)+\frac{1}{8}\text{Li}_3(u^2)\right)du
\end{align*}

Next, we use partial integration with $U=\left(\frac{7}{8}\zeta(3)u-\text{Li}_3(u)+\frac{1}{8}\text{Li}_3(u^2)\right)$ and the following relations:
\begin{align*}
\frac{d}{du}\text{Li}_3(u)&=\frac{d}{du}\sum_{n\geq 1}\frac{u^n}{n^3}=\sum_{n\geq 1}\frac{u^{n-1}}{n^2}=\frac{1}{u}\text{Li}_2(u)\\
\frac{d}{du}\text{Li}_3(u^2)&=\frac{d}{du}\sum_{n\geq 1}\frac{u^{2n}}{n^3}=2\sum_{n\geq 1}\frac{u^{2n-1}}{n^2}=\frac{2}{u}\text{Li}_2(u^2)\\
\text{Li}_2(u)+\text{Li}_2(-u)&=2\left(\sum_{n\geq 1}\frac{u^n}{n^2}+\sum_{n\geq 1}\frac{(-u)^n}{n^2}\right)=\frac{1}{2}\text{Li}_2(u^2)\\
\frac{d}{du}\text{Li}_2(u)&=\frac{d}{du}\sum_{n\geq 1}\frac{u^n}{n^2}=\sum_{n\geq 1}\frac{u^{n-1}}{n}=\frac{1}{u}\ln(1-u)\\
\frac{d}{du}\text{Li}_2(-u)&=\frac{d}{du}\sum_{n\geq 1}\frac{(-u)^n}{n^2}=-\sum_{n\geq 1}\frac{(-u)^{n-1}}{n}=\frac{1}{u}\ln(1+u)\\
\end{align*}

So we get
  \begin{align*}
\sum_{n\geq1}\frac{H_n}{(2n+1)^3}
&=2\left.\left(\frac{7}{8}\zeta(3)u-\text{Li}_3(u)+\frac{1}{8}\text{Li}_3(u^2)\right)\frac{1}{2}\ln\frac{1+u}{1-u}\right|_0^1\\
&-\int_0^1\left(\frac{7}{8}\zeta(3)-\frac{1}{u}\text{Li}_2(u)+\frac{1}{4u}\text{Li}_2(u^2)\right)\ln\frac{1+u}{1-u}du\\
&=-\int_0^1\left(\frac{7}{8}\zeta(3)-\frac{1}{u}\text{Li}_2(u)+\frac{1}{4u}\text{Li}_2(u^2)\right)\ln\frac{1+u}{1-u}du\\
&=-\frac{7}{8}\zeta(3)\int_0^1ln\frac{1+u}{1-u}du+\frac{1}{2}\int_0^1\left(\text{Li}_2(u)-\text{Li}_2(-u)\right)\frac{1}{u}\ln\frac{1+u}{1-u}du\\
&=-\frac{7}{8}\zeta(3)\left.\left(\ln(1-u^2)+u\ln\frac{1+u}{1-u}\right)\right|_0^1+\frac{1}{4}\left.\left(\text{Li}_2(u)-\text{Li}_2(-u)\right)^2\right|_0^1\\
&=-\frac{7}{4}\zeta(3)\ln(2)+\frac{9}{16}\zeta(2)^2
\end{align*}

Now, putting all together gives

\begin{align*}
\sum_{n\geq 1}(-1)^{n+1}\frac{H_{\lfloor n/2\rfloor}}{n^3}
&=-\frac{1}{8}\sum_{k\geq1}\frac{H_n}{n^3}+\sum_{n\geq1}\frac{H_n}{(2n+1)^3}\\
&=\frac{1}{2}\zeta(2)^2-\frac{7}{4}\zeta(3)\ln(2)
\end{align*}