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
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n} ={19\pi^{4} \over 1440} - {3 \over 4}\,\zeta\pars{3}\ln\pars{2}:\ {\large ?}. \qquad \zeta_{n}\pars{3} = \sum_{k = 1}^{n}{1 \over k^{3}}}$
\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}
$$\color{#66f}{\large% \sum_{n = 1}^{\infty}{\pars{-1}^{n - 1}\zeta_{n}\pars{3} \over n} ={19\pi^{4} \over 1440} - {3 \over 4}\,\zeta\pars{3}\ln\pars{2}} \approx 0.6604 $$
Let us recall the integral representation for a more general case that I introduced in a previous problem
$$ A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q} = \frac{\left( -1 \right) ^{q}}{\Gamma(q)}\int _{0}^{1}\!{\frac { \left( \ln\left( u \right) \right)^{q-1}{Li_{p}(-u)} }{ u\left( 1+ u \right) }}{du}.$$
where $Li_{p}(-u)$ is the polylogarithm function. Your sum can be written (see generalized harmonic numbers) as
$$ A(3,1)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(3)}_k}{k}= -\int _{0}^{1}\!{\frac { {Li_{3}(-u)} }{ u\left( 1+ u \right) }}{du}$$
$$ = \int _{0}^{1}\!{\frac { {Li_{3}(-u)} }{ \left( 1+ u \right) }}{du}-\int _{0}^{1}\!{\frac { {Li_{3}(-u)} }{ u }}{du} $$
$$ = \left(-\frac{3}{4}\,\ln \left( 2 \right) \zeta \left( 3 \right) +{\frac {1}{288}}\,{\pi }^{4}\right) - \left(- \frac{7}{720}\pi^4\right)$$
$$ \implies A(3,1) = \frac{19}{1440}\pi^4-\frac{3}{4}\,\ln \left( 2 \right) \zeta \left( 3 \right). $$
Note: To evaluate the integrals you can use integration by parts with $u={Li_{3}(-u)}$ or use computer algebra systems.