Different way of proving: $\sum_{n=1}^{\infty}\frac{\left(16(-1)^n+5\right)\left(\phi H_n+\frac{1}{n^3}\right)+\frac{11}{n^3}}{n^2}=\zeta(5)$

Question

$$\sum_{n=1}^{\infty}\frac{\left(16(-1)^n+5\right)\left(\phi H_n+\frac{1}{n^3}\right)+\frac{11}{n^3}}{n^2}=\zeta(5)\tag1$$

$\phi=\frac{1+\sqrt{5}}{2}$

$H_n=\sum_{k=1}^{n}\frac{1}{k}$

we expanded $(1)$ $$\sum_{n=1}^{\infty}\left(\frac{16(-1)^n\phi H_n}{n^2}+\frac{5\phi H_n}{n^2}+\frac{16(-1)^n}{n^5}+\frac{16}{n^5}\right)=\zeta(5)\tag2$$

$$\sum_{n=1}^{\infty}\left(\frac{16(-1)^n\phi H_n}{n^2}+\frac{5\phi H_n}{n^2}\right)-15\zeta(5)+15\zeta(5)=\zeta(5)\tag3$$

Is there another way of proving $(1)?$