Series Demystified 4

$S=\large \sum_{n=1}^{48} \frac {(n-1)!(n^3-n-1)}{n\sqrt{n^2(n+1)}+\sqrt{ n(n+1)^2}}$

$= \large \sum_{n=1}^{48} \frac {(n-1)!(n^3-n-1)(n\sqrt{n^2(n+1)}-\sqrt{ n(n+1)^2})}{(n\sqrt{n^2(n+1)}+\sqrt{ n(n+1)^2})(n\sqrt{n^2(n+1)}-\sqrt{ n(n+1)^2})}$

$= \large \sum_{n=1}^{48} \frac {(n-1)!(n^3-n-1)(n\sqrt{n^2(n+1)}-\sqrt{ n(n+1)^2})}{n^4(n+1)-n(n+1)^2}$

$= \large \sum_{n=1}^{48} \frac {(n-1)!(n^3-n-1)(n\sqrt{n^2(n+1)}-\sqrt{ n(n+1)^2})}{n(n+1)(n^3-n-1)}$

$=\large \sum_{n=1}^{48} \frac {(n-1)!(n^2\sqrt{n+1}-(n+1)\sqrt{ n})}{n(n+1)}$

$= \large \sum_{n=1}^{48} \frac {n!\sqrt{n+1} }{n+1} -\frac{(n-1)!\sqrt{ n}}{n}$

$S=\frac{48!\sqrt{49}}{49}-\frac{0!\sqrt{1}}{1}=\frac{48!}{7}-1$

$\implies \frac{48!}{S+1}=\boxed{7}$

Let us demystify the expression to be summated

$\dfrac{(n-1)![n^3 - n -1]}{n\sqrt{n^2(n+1)} + \sqrt{n(n+1)^2}}$

Taking $\sqrt{n(n+1)}$ common from the denominator

$\Rightarrow \dfrac{(n-1)![n^3 - n -1]}{\sqrt{n(n+1)}[n\sqrt{n} + \sqrt{n+1}]}$

$\Rightarrow \dfrac{(n-1)![n\sqrt{n} -\sqrt{n+1}]}{\sqrt{n(n+1)}}$

$\Rightarrow \dfrac{n(n-1)!\sqrt{n}}{\sqrt{n(n+1)}} - \dfrac{\sqrt{n+1}(n-1)!}{\sqrt{n(n+1)}}$

$\Rightarrow \dfrac{n!}{\sqrt{n+1}} - \dfrac{(n-1)!}{\sqrt{n}}$

This looks to be the $n^{th}$ term of the summation given.

$S = \displaystyle \sum_{n=1}^{48} \dfrac{n!}{\sqrt{n+1}} - \dfrac{(n-1)!}{\sqrt{n}}$

$S = \dfrac{1!}{\sqrt{2}} - \dfrac{0!}{\sqrt{1}} + \dfrac{2!}{\sqrt{3}} - \dfrac{1!}{\sqrt{2}} + \ldots + \dfrac{48!}{\sqrt{49}} - \dfrac{47!}{\sqrt{48}}$

$S = \dfrac{48!}{7} - 1$

Required answer is

$\dfrac{48!}{S+1} = \dfrac{48!}{\frac{48!}{7} - 1 + 1} = \dfrac{48!}{48!} 7 = \boxed{7}$