Haven't seen this before

The following identity is due to Lehmer (from a paper in the American Mathematical Monthly, 1985): $\sum_{m \ge 1} \frac{(2x)^{2m}}{m \binom{2m}{m}} = \frac{2x}{\sqrt{1-x^2}} \sin^{-1} x.$ Take the antiderivative of both sides (use integration by parts on the right, plug in 0 to get the integration constant); then plug in $k=m+1$ : $\begin{aligned} \sum_{m \ge 1} \frac{2^{2m} x^{2m+1} (m!)^2}{m(2m+1)(2m)!} &= 2x - 2(\sin^{-1} x)\sqrt{1-x^2} \\ \sum_{m \ge 1} \frac{2^{2m} x^{2m+1} (m-1)!m!}{(2m+1)!} &= 2x - 2(\sin^{-1} x)\sqrt{1-x^2} \\ \sum_{k \ge 0} \frac{2^{2k+2} x^{2k+3} k!(k+1)!}{(2k+3)!} &= 2x - 2(\sin^{-1} x)\sqrt{1-x^2} \\ \sum_{k \ge 0} \frac{2^{k+1} x^{2k+3} k! (2 \cdot 4 \cdots (2k+2))}{(2k+3)!} &= 2x - 2(\sin^{-1} x)\sqrt{1-x^2} \\ \sum_{k \ge 0} 2^{k+1} x^{2k+3} \frac{k!}{\prod_{j=0}^k (2j+3)} &= 2x - 2(\sin^{-1} x)\sqrt{1-x^2} \\ \end{aligned}$ Now plug in $x=1/\sqrt{2},$ to get $\begin{aligned} \frac1{\sqrt{2}} \sum_{k \ge 0} \frac{k!}{\prod_{j=0}^k (2j+3)} &= \frac2{\sqrt{2}} - \frac{2}{\sqrt{2}}\left(\sin^{-1} \frac1{\sqrt{2}} \right) \\ \sum_{k \ge 0} \frac{k!}{\prod_{j=0}^k (2j+3)} &= 2 - 2 \left(\frac{\pi}4\right) = 2-\frac{\pi}2. \end{aligned}$ So $a = 2, b = -1/2, c = 1$ and the answer is $\fbox{8875}.$

The following sum mentioned in the question can be written in the form $\sum _{ n=2 }^{ \infty }{ \frac { \left( n-2 \right) !{ 2 }^{ n }n! }{ \left( 2n \right) ! } } =\sum _{ n=2 }^{ \infty }{ \frac { \Gamma \left( n+1 \right) \Gamma \left( n \right)2^{n} }{ \Gamma \left( 2n+1 \right) \left( n-1 \right) } }$ we know that $\int _{ 0 }^{ 1 }{ { t }^{ x-1 }{ \left( 1-t \right) }^{ y-1 }dt=\frac { \Gamma \left( x \right) \Gamma \left( y \right) }{ \Gamma \left( x+y \right) } }$ Therefore the sum in terms of the gamma function can be written as $\sum _{ n=2 }^{ \infty }{ \frac { { 2 }^{ n } }{ n-1 } } \int _{ 0 }^{ 1 }{ { t }^{ n }{ \left( 1-t \right) }^{ n-1 }dt\; }$ on interchanging the sum and integral we get $\int _{ 0 }^{ 1 }{ \frac { 1 }{ 1-t } \sum _{ n=2 }^{ \infty }{ \frac { { \left( 2t\left( 1-t \right) \right) }^{ n } }{ n-1 } dt\; = } } \int _{ 0 }^{ 1 }{ -\left( 2t\ln { \left( 1-2t\left( 1-t \right) \right) } \right) dt }$ since $\boxed{{ \sum _{ n=2 }^{ \infty }{ \frac { { \left( 2t\left( 1-t \right) \right) }^{ n } }{ n-1 } = } }{ -2t\left( 1-t \right) \ln { \left( 1-2t\left( 1-t \right) \right) } }}$ by applying Integration by parts , we will get $\int _{ 0 }^{ 1 }{ -2t\ln { \left( 1-2t\left( 1-t \right) \right) } } dt=2-\frac { \pi }{ 2 }$ therefore $a=2$ , $b=-\frac{1}{2}$ and $c=1$ i.e $\boxed{10^{3}\left(8-\frac{1}{8}+1\right)=8875 }$

This solution will be heavily based on this paper by Renzo Sprugnoli.

Simplifying the summand gives us

$\frac { k! }{ \prod _{ j=0 }^{ k }{ \left( 2k+3 \right) } }=\frac { k! }{ \left( 2k+3 \right) !! } =\frac { { 2 }^{ k+1 }k!(k+1)! }{ \left( 2k+3 \right) ! } =\frac { { 2 }^{ k+1 } }{ \left( 2k+3 \right) \left( k+1 \right) \left( \begin{matrix} 2k+2 \\ k+1 \end{matrix} \right) }$

After reindexing we get

$\sum _{ k=0 }^{ \infty }{ \frac { { 2 }^{ k+1 } }{ \left( 2k+3 \right) \left( k+1 \right) \left( \begin{matrix} 2k+2 \\ k+1 \end{matrix} \right) } } =\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ \left( 2k+1 \right) \left( k \right) \left( \begin{matrix} 2k \\ k \end{matrix} \right) } } =\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } } -\frac { { 2 }^{ k+1 } }{ \left( 2k+1 \right) \left( \begin{matrix} 2k \\ k \end{matrix} \right) }$

We can further simplify the term on the right to get

$\sum _{ k=1 }^{ \infty }{\frac { { 2 }^{ k+1 } }{ \left( 2k+1 \right) \left( \begin{matrix} 2k \\ k \end{matrix} \right) }}=\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k+1 } }{ \frac { { \left( k+1 \right) }^{ 2 } }{ 2\left( k+1 \right) } \left( \begin{matrix} 2k+2 \\ k+1 \end{matrix} \right) } } =\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k+2 } }{ \left( k+1 \right) \left( \begin{matrix} 2k+2 \\ k+1 \end{matrix} \right) } } =2\left( \sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } } -1 \right)$ $\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } } -\frac { { 2 }^{ k+1 } }{ \left( 2k+1 \right) \left( \begin{matrix} 2k \\ k \end{matrix} \right) } =\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } } -2\left( \sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } } -1 \right)=2-\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } }$

Now, let ${Z}_{k}$ denote $\frac { { 2 }^{ k } }{ \left( \begin{matrix} 2k \\ k \end{matrix} \right) }$ , and its generating function be $Z\left( t \right) =\sum _{ k=0 }^{ \infty }{ { Z }_{ k } } { t }^{ k }$ . We can show that ${ Z }_{ k+1 }=\frac { { 2 }^{ k+1 } }{ \left( \begin{matrix} 2k+2 \\ k+1 \end{matrix} \right) } =\frac { 2 }{ \frac { 2\left( k+1 \right) \left( 2k+1 \right) }{ { \left( k+1 \right) }^{ 2 } } } \frac { { 2 }^{ k } }{ \left( \begin{matrix} 2k \\ k \end{matrix} \right) } =\frac { k+1 }{ 2k+1 } \frac { { 2 }^{ k } }{ \left( \begin{matrix} 2k \\ k \end{matrix} \right) } =\frac { k+1 }{ 2k+1 } { Z }_{ k }$ $2\left( k+1 \right) { Z }_{ k+1 }-{ Z }_{ k+1 }=k{ Z }_{ k }+{ Z }_{ k }$ $2{ Z'\left( t \right) }-\frac { Z\left( t \right) -{ Z }_{ 0 } }{ t } =tZ'\left( t \right) +Z\left( t \right)$ $t\left( 2-t \right) { Z'\left( t \right) }-\left( 1+t \right) Z\left( t \right) +1=0$

This differential equation can be solved by integrating factor to yield $Z(t)=\frac { 2 }{ 2-t } +\frac { 2\sqrt { t } }{ { \left( 2-t \right) }^{ \frac { 3 }{ 2 } } } \arcsin { \sqrt { \frac { t }{ 2 } } }$

Now, let $W(t)=\sum _{ k=1 }^{ \infty }{ { W }_{ k }{ t }^{ k } } =\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } { t }^{ k } }$ . To get $W(t)$ from $Z(t)$ , we can observe that ${W}_{k}=\frac { { Z }_{ k } }{ k }$ . Knowing this, it is simply a matter of reindexing then integrating. We can thus define $W(t)=\int _{ 0 }^{ t }{ \frac { Z\left( z \right) -{ Z }_{ 0 } }{ z } } dz=\int _{ 0 }^{ t }{ \frac { \left( \frac { 2 }{ 2-z } +\frac { 2\sqrt { z } }{ { \left( 2-z \right) }^{ \frac { 3 }{ 2 } } } \arcsin { \sqrt { \frac { z }{ 2 } } } \right) -1 }{ z } } dz=2\sqrt { \frac { t }{ 2-t } } \arcsin { \sqrt { \frac { t }{ 2 } } }$

Going back to the original sum in question, it is obvious that $W(1)=\sum _{ k=1 }^{ \infty }{ \frac { { 2 }^{ k } }{ k\left( \begin{matrix} 2k \\ k \end{matrix} \right) } }=2\sqrt { \frac { 1 }{ 2-1 } } \arcsin { \sqrt { \frac { 1 }{ 2 } } } =2\arcsin { \frac { 1 }{ \sqrt { 2 } } } =\frac { \pi }{ 2 }$

Hence $\sum _{ k=0 }^{ \infty }{ \frac { k! }{ \prod _{ j=0 }^{ k }{ \left( 2j+3 \right) } } } =\boxed{2-\frac { \pi }{ 2 }}$

There is a really beautiful approach to solve this problem. Here, I refer to the $\bold{wallis \space product}$ for odd powers of $\sin(x)$ (which you can pretty easily derive using the reduction formula), which is,

$\int_0^{\frac{\pi}{2}}\sin^{2n+1}(x) dx = \frac{(2^n)n!}{(1)(3)(5)\dots(2n+1)} \space \forall \space n \space \in \space \mathbb N$ .

Using this, the above sum can be written as $\sum_{k=1}^{\infty}\frac{1}{(2^k)(k)}\int_0^{\frac{\pi}{2}}\sin^{2k+1}(x) \space dx$

Swapping the integral and summation,

$\int_0^{\frac{\pi}{2}}\sin(x)\sum_{k=1}^{\infty} \frac{(\frac{\sin^2(x)}{2})^k}{k} \space dx$

If you take $\frac{\sin^2(x)}{2} = t$ , you note that the term $\frac{(\frac{\sin^2(x)}{2})^k}{k} = \int t^{k-1} dt$ .

The sum as of now is:

$\int_0^{\frac{\pi}{2}}\sin(x)\sum_{k=1}^{\infty} \int t^{k-1} dt \space dx$

Swapping the inner integral and summation,

$\int_0^{\frac{\pi}{2}}\sin(x) \int \sum_{k=1}^{\infty} t^{k-1} dt \space dx$

Because $0 < t < 1 \space \forall \space x \space \in \space \mathbb R$ ,

$\int_0^{\frac{\pi}{2}}\sin(x) \int 1/(1-t) dt \space dx$

After a couple of steps, this integral is:

$\int_0^{\frac{\pi}{2}}\sin(x) \ln{|1 + \cos^2(x)|} \space dx$ .

This can be easily solved using u-substitution to give the result $2 - \frac{\pi}{2}$

Let us denote the product $P= \prod_{j=0}^{k} (2j+3)=\prod_{j=1}^{k+1} (2j+1)=(2k+3)!!$ The latter double factorial can further be expressed in terms of gamma function as $(2k+3)!!=\dfrac{2^{k+2}}{\sqrt{\pi}}\Gamma\left(k+\dfrac{5}{2}\right)=\dfrac{2^{k+2}}{\sqrt{\pi}}\left(\dfrac{(2k+4)!\sqrt{\pi}}{2^{2k+4} (n+2)!}\right)$ and simplifying the latter expression we can get $\sum_{k=0}^{\infty}k !P^{-1} =\sum_{k=0}^{\infty} \dfrac{2^k(k!)^2}{(2k+1)(2k+3)(2k)!}=\dfrac{1}{2}\sum_{k=0}^{\infty} \left(\dfrac{1}{2k+1}-\dfrac{1}{2 k+3}\right)2^n\binom{2k}{k}^{-1}$ Since $2(\sin^{-1}x)^2=\sum_{k=1}^{\infty}\dfrac{(2x)^{2k}}{k^2}\binom{2k}{k}^{-1}$ we want to get rid of $k^2$ term in the denominator as first we differentiate the function and we repeate once again and hence on multiplying by $x$ we can get $\begin{aligned} \dfrac{2\sin^{-1} x}{\sqrt{1-x^2}} & =& \sum_{k=1}^{\infty} \dfrac{(2k)^{2k}}{k}\binom{2k}{k}^{-1}\\2 x\dfrac{d}{dx}\dfrac{\sin^{-1}x}{\sqrt{1-x^2}} = \dfrac{x\sin^{-1}x+x^2\sqrt{1-x^2}}{(1-x^2)\sqrt{1-x^2}}&=&\sum_{k=1}^{\infty}(2x)^{2k}\binom{2k}{k}^{-1}\end{aligned}$ now we integrate with respect to x from $0$ to $\frac{1}{\sqrt{2}}$ we get $\dfrac{1}{\sqrt 2}\sum_{k=1}^{\infty}\dfrac{2^k}{2k+1}\binom{2k}{k}^{-1} =\dfrac{\sin^{-1}x}{\sqrt{1-x^2}} -x{\LARGE|}_0^{\frac{1}{\sqrt 2}}=\dfrac{1}{\sqrt2}\left(\dfrac{\pi}{2}-1\right)$ and hence we have $\sum_{k=0}^{\infty} \dfrac{2^k}{2k+1}\binom{2k}{k}^{-1}=\dfrac{\pi}{2}$ similarly we multiply by $x^2$ and hnece on integrating from to $0$ to $\frac{1}{\sqrt 2}$ . We get $\begin{aligned} \sum_{k=1}^{\infty} \dfrac{2^k}{2k+3} \binom{2k}{k}^{-1}& = 2\sqrt2\left(\dfrac{3\pi}{4\sqrt 2}-\dfrac{13}{6\sqrt2}\right)\\ \sum_{k=0}^{\infty}\dfrac{2^k}{2k+3}\binom{2k}{k}^{-1}& =\dfrac{3\pi}{2}-4\end{aligned}$ and hence $\sum_{k=0}^{\infty}k!\left(\prod_{j=0}^{k} (2j+3)\right)^{-1}=\dfrac{1}{2}\left(\dfrac{\pi}{2}+4-\dfrac{3\pi}{2}\right)=2-\dfrac{\pi}{2}$

I left the working of integral since its not difficult to do. One can now easily derive the following more identities $\begin{aligned} \sum_{k\geq 0}\dfrac{2^k}{2k+1}\binom{2k}{k}^{-1} &=\frac{\pi}{2}\\ \sum_{k\geq 0}\dfrac{2^k}{2k+3}\binom{2k}{k}^{-1} & =\dfrac{3\pi}{2}-4\\\sum_{k\geq 0}\dfrac{2^k}{2k+5}\binom{2k}{k}^{-1} &=\dfrac{23\pi}{6}-\dfrac{104}{9}\\ \sum_{k\geq 0}\dfrac{2^k}{2k+7}\binom{2k}{k}^{-1} &=\dfrac{91\pi}{10}-\dfrac{2116}{75}\\\sum_{k\geq 0}\dfrac{2^k}{2k+9}\binom{2k}{k}^{-1} &=\dfrac{1451\pi}{70}-\dfrac{238912}{3675} \\ \sum_{k\geq 0}\dfrac{2^k}{2k+11}\binom{2k}{k}^{-1} & =\dfrac{5797\pi}{126}-\dfrac{2863204}{19845}\end{aligned}$ and soon. We can even solve using beta function.