Find the values of a, b, c

Question

Given $$\sum_{k=0}^{\infty}\frac{k!}{\Pi_{j=0}^{k}(2j+3)} = a + b\pi^c$$ where $a$, $c$ - integers, $b$ - is a rational number. Find a, b, c.

Please help or give me a tip on how to approach this kind of problem. Thank you.

Answer 1

$$S=\sum_{k=0}^\infty \frac{k!}{\prod_{n=0}^k (2n+3)}$$ $$\prod_{n=0}^k (2n+3)=2^k\prod_{n=0}^k \left(n+\frac{3}{2}\right)=2^k\frac{\Gamma(k+1+\frac{3}{2})}{\Gamma(\frac{3}{2})}$$

$$S=\sum_{k=0}^\infty \frac{\Gamma(k+1)\Gamma(\frac{3}{2})}{2^k\Gamma(k+\frac{5}{2})}=\sum_{k=0}^\infty \frac{\mathfrak{B}(k+1,\frac{3}{2})}{2^k}=\int_0^1\sum_{k=0}^\infty \left(\frac{t}{2}\right)^k(1-t)^{\frac{1}{2}}dt$$

$$S=\int_0^1\frac{(1-t)^{\frac{1}{2}}}{1-\frac{t}{2}}dt=2\int_0^1\underbrace{\frac{(1-t)^{\frac{1}{2}}}{2-t}}_{t=1-z}dt=2\int_0^1\underbrace{\frac{z^{\frac{1}{2}}}{1+z}}_{z^{\frac{1}{2}}=y}dz$$ $$S=4\int_0^1\frac{y^2}{1+y^2}dy=4-\pi$$