Til the last breath

Inspiration

If you found this problem easy do try this one too.. :-) $\large\mathfrak{S}=\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{n(n+2)}{\displaystyle\sum_{r=1}^n r(r+1)(r+2)}\right)$ Using formula for sum of n,n $^2$ ,n $^3$ denominator can be simplified as: $\large \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{\cancel{n(n+2)}}{\frac{\cancel{ n}(n+1)\cancel{(n+2)}(n+3)}{4}}\right)$ $\large=\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{4}{(n+1)(n+3)}\right)$ Using Partial Fractions-Cover Up rule :- $\dfrac{4}{(n+1)(n+3)}=2\left(\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)$ Therefore, $\large\mathfrak S=2\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)$ $\large\color{#69047E}{\mathbf{A Telescopic Series}}$ $\large =2\left(\dfrac{1}{2}+\dfrac{1}{3}\right)$ $\large =\dfrac{5}{3}$ $\Large\implies \color{#456461}{\phi}=5\times 3=15$

$\Large\color{#69047E}{\varphi}=\color{#456461}{\phi}^{\color{#EC7300}{\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}\cdots}}$ $~~~~~~~~~~~\Large\implies \color{#69047E}{\varphi}=\color{#456461}{\phi}^{\color{#EC7300}{2}}~~~~~~(**)~~~~~~~$ $\large\therefore \color{#3D99F6}{\left(\dfrac{\sqrt{\color{#69047E}{\varphi}}}{5}\right)!}=\left(\dfrac{\color{#456461}{\phi}}{5}\right)!=3!=\boxed 6$

$\Large(**)\small{\mathbf{( AGP)}}$ $\small{\mathfrak{S}=\frac { 1 }{ 2} +\frac { 2 }{ 2^2 } +\frac { 3 }{ 2^3} +\frac { 4 }{ 2^4 } +\dots~~~~~~}$ $\small{\dfrac{1}{2}\mathfrak{S}=~~~~~~~~\frac { 1 }{2^2 } +\frac { 2 }{ 2^3} +\frac { 3 }{ 2^4 } +\frac { 4 }{ 2^5} +\dots}$ Subtracting we get: $\small{\dfrac{1}{2}\mathfrak{S}=\frac { 1 }{ 2 }+ \frac { 1 }{2^2} +\frac { 1}{ 2^3} +\frac { 1 }{ 2^4 } +\frac { 1 }{ 2^5} +\dots}$ $\small{\mathbf{(Infinite ~GP)}}$ $\small{\Rightarrow \mathfrak{S}=2\left(\dfrac{\frac{1}{2}}{1-\frac{1}{2}}\right)=2}$ $\small{\therefore \color{#EC7300}{\mathfrak S=2}}$

This is very cool. Glad i could solve it easily.