この記事では、
$$\displaystyle\sum_{a_1,\ldots,a_n>0}\frac1{\displaystyle\prod_{k=1}^na_k\left(\displaystyle\sum_{k=1}^n a_k-1 \right)}=\zeta(n)\Gamma(n+1)$$
この一般化の証明を書きます。
$ \begin{align} &\sum_{a_1,\ldots,a_n>0}\frac1{\displaystyle\prod_{k=1}^na_k\left(\displaystyle\sum_{k=1}^n a_k-1 \right)}\\ =&\sum_{a_1,\ldots,a_n>0}\frac1{\displaystyle\prod_{k=1}^na_k}\int_0^1x^{\displaystyle\Sigma_{k=1}^na_k-2}dx\\ =&\int_0^1 \frac1{x^2}\sum_{a_1,\ldots,a_n>0}\frac{x^{\displaystyle\Sigma_{k=1}^na_k}}{\displaystyle\prod_{k=1}^na_k}dx\\ =&\int_0^1 \frac{(-\log(1-x))^n}{x^2}dx\\ =&\int_0^1 (-\log x)^n\sum_{k=1}^\infty kx^{k-1} dx\\ =&\sum_{k=1}^\infty k\int_0^\infty t^ne^{-kt}dt\\ =&\zeta(n)\Gamma(n+1)\\ \end{align} $
よって、$\displaystyle\sum_{a_1,\ldots,a_n>0}\frac1{\displaystyle\prod_{k=1}^na_k\left(\displaystyle\sum_{k=1}^n a_k-1 \right)}=\zeta(n)\Gamma(n+1)$が示されました。□