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 
 A: 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.  
