Search for Beta

$\displaystyle \sum_{n=0}^{\infty}\dfrac{H_{n}}{(n+1)^{4}} = \sum_{n=1}^{\infty}\left(\dfrac{H_{n}}{n^{4}}\right) - \zeta(5)$

Now we will evaluate $\sum_{n=1}^{\infty}\dfrac{H_{n}}{n^{4}}$

Consider $\displaystyle \dfrac{1}{n^{4}} = -6\int_{0}^{1} x^{n-1}\ln ^{3} dx$

Multiplying both sides with $H_{n}$ ,

$\displaystyle \dfrac{H_{n}}{n^{4}} = -6\int_{0}^{1} (H_{n}x^{n-1})\ln ^{3} dx$

$\implies \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{4}} = -6\int_{0}^{1} \sum_{n=1}^{\infty}(H_{n}x^{n})\dfrac{\ln ^{3}}{x} dx$

Using generating function for the harmonic numbers $\rightarrow \sum_{n=1}^{\infty} H_{n} x^{n} = \dfrac{-ln(1-x)}{1-x}$ ,

$\implies \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{4}} = 6\int_{0}^{1}\dfrac{\ln(1-x)\ln ^{3}}{x(1-x)} dx$

The integral is easy to evaluate, it is just derivative of beta function.Let $K$ denote that integral.

Then $\displaystyle K = 6\dfrac{\partial^{4}F}{\partial a \partial b^3}$ , where $F = \int_{0}^{1} x^{b}(1-x)^{a}dx$

It evaluates to $3\zeta(5)-\zeta(3)\zeta(2)$

Thus $\displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^{4}} -\zeta(5) = 2\zeta(5)-\zeta(3)\zeta(2)$

Thus final answer $=2+5+3+2=\boxed{\boxed{12}}$