$\underline{\text{Motivation}:-}$
Recently I saw this video from Micheal Penn. After seeing the entire video I realized that all the things he did can be easily generalized.
In that video he showed,
$\textstyle\displaystyle{\sum_{m,n\ge 1}\frac{1}{mn(m+n)}=2\zeta(3)}$
So this is my motivation.
$\underline{\text{My work}:-}$
$■Claim:-$
$\textstyle\displaystyle{\sum_{a_1,\dots, a_n\geq 1}\frac{1}{\prod_{i=1}^{n}a_i\sum_{i=1}^{n}a_i}}=n!\zeta(n+1)$
$■Proof:-$
$\textstyle\displaystyle{\sum_{a_1,\dots, a_n\geq 1}\frac{1}{\prod_{i=1}^{n}a_i\sum_{i=1}^{n}a_i}}$
$=\sum_{a_1=1}^{\infty}\cdots\sum_{a_n=1}^{\infty}\frac{1}{a_1\cdots a_n(a_1+\cdots+a_n)}$
Now we use the fact that $\textstyle\displaystyle{\frac{1}{n}=\int_{0}^{1}x^{n-1}dx}$
So we get,
$\sum_{a_1=1}^{\infty}\cdots\sum_{a_n=1}^{\infty}\left(\int_{0}^{1}x_1^{a_1-1}dx_1\cdots\int_{0}^{1}x_n^{a_n-1}dx_n\int_{0}^{1}y^{a_1+\cdots+a_n-1}dy\right)$
$=\int\cdots\int_{[0,1]^{n+1}}y^{n-1}\left(\sum_{a_1=1}^{\infty}x_1^{a_1-1}y^{a_1-1}\cdots\sum_{a_n=1}^{\infty}x_n^{a_n-1}y^{a_n-1}\right)dx_1\cdots dx_ndy$
Next we use the fact that $\sum_{n=1}^{\infty}x^{n-1}=\frac{1}{1-x}$
Giving us,
$\int\cdots\int_{[0,1]^{n+1}}\frac{y^{n-1}}{(1-x_1y)\cdots(1-x_ny)}dx_1\cdots dx_ndy$
$=\int_{0}^{1}y^{n-1}\prod_{k=1}^{n}\int_{0}^{1}\frac{1}{1-x_1y}dx_kdy$
$=\int_{0}^{1}y^{n-1}\prod_{k=1}^{n}(-\frac{\ln(1-x_ky)}{y}\bigg|_{0}^{1})dy$
$=(-1)^n\int_{0}^{1}\frac{\ln^n(1-y)}{y}dy$
Substituting $z=1-y$ gives us,
$(-1)^n\int_{0}^{1}\frac{\ln^n(z)}{1-z}dz$
$=(-1)^n\sum_{k=1}^{\infty}\int_{0}^{1}z^k\ln^n(z)dz$
Substituting $u=-\ln(z)$ gives us,
$\sum_{k=1}^{\infty}\int_{0}^{\infty}u^ne^{-(k+1)u}du$
Finally substituting $t=(k+1)u$ gives us,
$\sum_{k=0}^{\infty}\frac{1}{(k+1)^{n+1}}\int_{0}^{\infty}t^ne^{-t}dt$
$=n!\zeta(n+1)$
$\therefore\textstyle\displaystyle{\sum_{a_1,\dots, a_n\geq 1}\frac{1}{\prod_{i=1}^{n}a_i\sum_{i=1}^{n}a_i}}=n!\zeta(n+1)$
There is another way, if we take
$(-1)^n\int_{0}^{1}\frac{\ln^n(z)}{1-z}dz$
And instead turning it into a sum. We just substitute $u=-\ln(z)$ then we get,
$\int_{0}^{\infty}\frac{u^n}{e^u-1}du=\Gamma(n+1)\zeta(n+1)$
And since $n\in\mathbb{Z}^{+}$ we can write it as
$n!\zeta(n+1)$
$\underline{\text{My Question}:-}$
I would like to have my proof reviewed. But I am really asking for better ways to derive this.
