Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$ Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$

A related problem. Here is another approach. Recalling the generating function of the harmonic numbers
$$ \sum_{n=1}^{\infty} H_n x^n = \frac{\ln(1-x)}{x-1} \implies (xD)^4\sum_{n=1}^{\infty} H_n x^n=(xD)^4 \frac{\ln(1-x)}{x-1}, $$
$$ \implies \sum_{n=1}^{\infty} H_n n^4 x^n = {\frac {x \left( 1+11\,x+11\,{x}^{2}+{x}^{3} \right) \ln  \left( 1-x
 \right) }{ \left( -1+x \right) ^{5}}}+{\frac {x \left( -1-27\,{x}^{2}
-18\,x-4\,{x}^{3} \right) }{ \left( -1+x \right) ^{5}}}.$$
Substituting $x=-\frac{1}{2}$ in the above identity gives the desired result
$$ \sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}={\frac {28}{243}}-{\frac {10}{81}}\,\ln  \left( \frac{3}{2} \right).  $$ 
