Hyper Series

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {(n+2)!(n+3)!}{n!(n+10)!} \\ & = \sum_{n=0}^\infty \frac {(n+1)(n+2)}{(n+4)(n+5)(n+6)(n+7)(n+8)(n+9)(n+10)} & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac 1{120} \sum_{n=0}^\infty \left(\frac 1{n+4} - \frac {12}{n+5} + \frac {50}{n+6} - \frac {100}{n+7} + \frac {105}{n+8} - \frac {56}{n+9} + \frac {12}{n+10} \right) \end{aligned}$

We note that the sum of nominators $1-12+50-100+105-56+12 = 0$ . Therefore,

$\begin{aligned} S & = \frac 1{120} \left(\frac 14 + \frac {1-12}5 + \frac {1-12+50}6 + \frac {1-12+50-100}7 + \frac {1-12+50-100+105}8 + \frac {1-12+50-100+105-56}9 \right) \\ & = \frac 1{120} \left(\frac 14 - \frac {11}5 + \frac {39}6 - \frac {61}7 + \frac {44}8 - \frac {12}9 \right) \\ & = \frac 1{120} \times \frac 1{420} \\ & = \boxed {\dfrac 1{50400}} \end{aligned}$

We can write the given series as $\sum _{ n=0 }^{ \infty }{ \frac { (n+3-1)!(n+4-1)! }{ (n+11-1)!n! } } =\ x$ , where $x$ is considered to be its value for the time being ,we know that $\Gamma (s)=(s-1)!$ therefore the series can also be written as $\sum _{ n=0 }^{ \infty }{ \frac { \Gamma (n+3)\Gamma (n+4) }{ \Gamma (n+11)n! } } =x$ , we"ll multiply the the LHS and the RHS by $\frac { \Gamma (11) }{ \Gamma (3)\Gamma (4) }$ . i.e $\frac { \Gamma (11) }{ \Gamma (3)\Gamma (4) } \sum _{ n=0 }^{ \infty }{ \frac { \Gamma (n+3)\Gamma (n+4) }{ \Gamma (n+11)n! } } =\frac { \Gamma (11) }{ \Gamma (3)\Gamma (4) } \ x$ . we know that ${ (s) }_{ n }=\frac { \Gamma (s+n) }{ \Gamma (s) }$ where ${ (s })_{ n }\ =\ s(s+1)(s+2)....(s+n-1)$ is a pochammer notation for rising factorial, so the series can be written as $\sum _{ n=0 }^{ \infty }{ \frac { { (3) }_{ n }{ (4) }_{ n } }{ { (11) }_{ n }n! } } =\frac { \Gamma (11) }{ \Gamma (3)\Gamma (4) } x$ The series written on the LHS is a gauss hypergeometric series which is of the form ${ _{ 2 }{ F }_{ 1 }\left( \alpha ,\beta ;\gamma ;1 \right) }=\sum _{ n=0 }^{ \infty }{ \frac { { (\alpha ) }_{ n }{ (\beta ) }_{ n }{ (1) }^{ n } }{ { (\gamma ) }_{ n }(n)! } }$
if $\Re e\quad (\gamma -\alpha -\beta ) > 0$ the result for this series is $\sum _{ n=0 }^{ \infty }{ \frac { { (\alpha ) }_{ n }{ (\beta ) }_{ n }{ (1) }^{ n } }{ { (\gamma ) }_{ n }(n)! } =\frac { \Gamma (\gamma )\Gamma (\gamma -\alpha -\beta ) }{ \Gamma (\gamma -\alpha )\Gamma (\gamma -\beta ) } }$ we came to know that $\gamma = 11$ , $\alpha = 3$ and $\beta= 4$ therefore we get $\sum _{ n=0 }^{ \infty }{ \frac { { (3) }_{ n }{ (4) }_{ n }{ (1) }^{ n } }{ { (11) }_{ n }(n)! } =\frac { \Gamma (3)\Gamma (11-3-4) }{ \Gamma (11-3)\Gamma (11-4) } } =\frac { \Gamma (11)\Gamma (4) }{ \Gamma (8)\Gamma (7) }$ we'll substitute the value of the hypergeometric series on the 4th equation of this solution part therefore $\frac { \Gamma (11)\Gamma (4) }{ \Gamma (8)\Gamma (7) } =\quad x{ \frac { \Gamma (11) }{ \Gamma (3)\Gamma (4) } }$ on solving we"ll get $x =\frac { \Gamma (3)\Gamma (4)\Gamma (4) }{ \Gamma (8)\Gamma (7) } =\frac { (2)!(3)!(3)! }{ (7)!(6)! } =\frac { 1 }{ 50400 }$ so the value of $a$ as per the question is equal to $50400$ i.e $\boxed{a = 50400}$