summation - How find this $sum_{n=1}^{infty}frac{(-1)^{n-1}zeta_{n}(3)}{n}=?$

Question:

show that
$$\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}\zeta_{n}(3)}{n}=\dfrac{19\pi^4}{1440}-\dfrac{3}{4}\zeta{(3)}\ln{2}?$$

where $$\zeta_{n}(3)=\sum_{k=1}^{n}\dfrac{1}{k^3}$$

But I use this computer find this

and my reslut is wrong? Thank you

Answer

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$\ds{\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n}
={19\pi^{4} \over 1440} - {3 \over 4}\,\zeta\pars{3}\ln\pars{2}:\ {\large ?}.
\qquad \zeta_{n}\pars{3} = \sum_{k = 1}^{n}{1 \over k^{3}}}$

\begin{align}&\color{#c00000}{\sum_{n = 1}^{\infty}
{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n}}
=\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1} \over n}\sum_{k = 1}^{n}{1 \over k^{3}}
=\sum_{k = 1}^{\infty}{1 \over k^{3}}
\color{#00f}{\sum_{n = k}^{\infty}{\pars{-1}^{n - 1} \over n}}
\end{align}

\begin{align}&\color{#00f}{\sum_{n = k}^{\infty}

{\pars{-1}^{n - 1} \over n}}\color{#00f}
=\sum_{n = k}^{\infty}\pars{-1}^{n - 1}\int_{0}^{1}x^{n - 1}\,\dd x
=\int_{0}^{1}\sum_{n = k}^{\infty}\pars{-x}^{n - 1}\,\dd x
=\int_{0}^{1}{\pars{-x}^{k - 1} \over 1 - \pars{-x}}\,\dd x
\\[3mm]&=\int_{0}^{1}{\pars{-x}^{k - 1} \over 1 + x}\,\dd x
\end{align}

\begin{align}&\color{#c00000}{\sum_{n = 1}^{\infty}
{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n}}

=\sum_{k = 1}^{\infty}{1 \over k^{3}}
\int_{0}^{1}{\pars{-x}^{k - 1} \over 1 + x}\,\dd x
=-\int_{0}^{1}\sum_{k = 1}^{\infty}{\pars{-x}^{k} \over k^{3}}
\,{1 \over x\pars{1 + x}}\,\dd x
\\[3mm]&=-\int_{0}^{1}{{\rm Li}_{3}\pars{-x} \over x\pars{1 + x}}\,\dd x
=\int_{-1}^{0}{{\rm Li}_{3}\pars{x} \over x\pars{1 - x}}\,\dd x
=\int_{-1}^{0}{{\rm Li}_{3}\pars{x} \over x}\,\dd x
+\int_{-1}^{0}{{\rm Li}_{3}\pars{x} \over 1 - x}\,\dd x
\\[3mm]&=-{\rm Li}_{4}\pars{-1} + {\rm Li}_{3}\pars{-1}\ln\pars{2}
+\int_{-1}^{0}\ln\pars{1 - x}{\rm Li}_{3}'\pars{x}\,\dd x

\\[3mm]&=-{\rm Li}_{4}\pars{-1} + {\rm Li}_{3}\pars{-1}\ln\pars{2}
-\int_{-1}^{0}x{\rm Li}_{2}'\pars{x}\,{{\rm Li}_{2}\pars{x} \over x}\,\dd x
\end{align}

\begin{align}
\color{#c00000}{\sum_{n = 1}^{\infty}
{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n}}
&={\large-{\rm Li}_{4}\pars{-1} + {\rm Li}_{3}\pars{-1}\ln\pars{2}
+\half\,{\rm Li}_{2}^{2}\pars{-1}}

\\[3mm]\mbox{and}&\qquad
\left\lbrace\begin{array}{rcl}
{\rm Li}_{4}\pars{-1} & = & -\,{7\pi^{4} \over 720}
\\
{\rm Li}_{3}\pars{-1} & = & -\,{3 \over 4}\,\zeta\pars{3}
\\
{\rm Li}_{2}\pars{-1} & = & -\,{\pi^{2} \over 12}
\end{array}\right.
\end{align}

$$\color{#66f}{\large%
\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n}
={19\pi^{4} \over 1440} - {3 \over 4}\,\zeta\pars{3}\ln\pars{2}}
\approx 0.6604
$$