Too much to solve!

$Let~\Large \mathfrak{C}=\displaystyle \sum_{x=0}^{\infty} \dfrac{1}{\displaystyle \prod_{n=1}^{m+1} (x+n)}$ $\large =\frac{1}{m}\cdot\sum_{x=0}^{\infty}(\frac{1}{\displaystyle\prod_{n=1}^{m}(n+x)}-\frac{1}{\displaystyle\prod_{n=2}^{m+1}(n+x)})$

$\large\color{#D61F06}{\mathcal{\text{(A TELESCOPIC SERIES)}}}$

$\large=\frac{1}{m}(\frac{1}{\displaystyle\prod_{n=1}^{m}(n)})$

$\Large\color{#3D99F6}{=\dfrac{1}{m\times m!}}$ $\large \therefore \displaystyle \sum_{m=1}^{26} \left [\sum_{x=0}^\infty \prod_{n=1}^{m+1} \dfrac1{x+n} \right ]^{-1}$ $= \displaystyle \sum_{m=1}^{26}m\times m!$ $\large = \displaystyle \sum_{m=1}^{26} ((m+1)!- m! )$ $\Large =27!-1$ $\huge \therefore \sqrt[3]{\omega}= \sqrt[3]{27}=\boxed{\color{#007fff}{3}}$

$S = \displaystyle \sum_{m=1}^{26} \displaystyle \left[\sum_{x=0}^{\infty} \prod_{n=1}^{m+1} \dfrac{1}{x+n} \right]^{-1}$
Consider
$\displaystyle \left[ \sum_{x=0}^{\infty} \prod_{n=1}^{m+1} \dfrac{1}{x+n} \right] = \displaystyle\sum_{x=0}^{\infty} \dfrac{1}{(x+1)(x+2)(x+3)\ldots(x+m+1)} = \dfrac{1}{m} \times \left[\dfrac{1}{(x+1)(x+2)(x+3)\ldots(x+m)} - \dfrac{1}{(x+2)(x+3)(x+4)\ldots(x+m+1)} \right]$
This forms a telescopic series, i.e a series where successive terms cancel out each other and we are left with,

$\displaystyle \left[ \sum_{x=0}^{\infty} \prod_{n=1}^{m+1} \dfrac{1}{x+n} \right ] = \dfrac{1}{m} \times \left[\dfrac{1}{m!} - \displaystyle \lim_{x \rightarrow \infty} \dfrac{1}{(x+2)(x+3)(x+4)\ldots(x+m+1)}\right] = \dfrac{1}{m\cdot m!}$
$S = \displaystyle \sum_{m=1}^{26} \left[\dfrac{1}{m \cdot m!}\right]^{-1} = \displaystyle \sum_{m=1}^{26}m\cdot m! = \sum_{m=1}^{26} (m+1-1)\cdot m! = \sum_{m=1}^{26} (m+1)! - m!$
This forms another telecsopic series where the terms cancel out as follows,

$\therefore S = 2! - 1! + 3! - 2! + \ldots 27! - 26! = 27! - 1$
$\therefore S = 27! -1$
Comparing we get $\omega = 27$
$\sqrt[3]{\omega} = \sqrt[3]{27} = 3$