Why is $\sum_{n\geq 0}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}=\frac{1}{p}\sum_{n\geq 0}\binom{-1/2}{n}\frac{1}{p^{2n}} $ true?

Question

Given the following working for showing that $$\int_0^\infty J_0(ax)e^{-px} dx=\frac{1}{\sqrt{a^2+p^2}}$$ (without using Laplace transform tables):

$$ J_0(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{4^n n!^2} $$ and since for any $p>0$ $$\int_{0}^{+\infty}x^{2n}e^{-px}\,dx = \frac{(2n)!}{p^{2n+1}}$$ we formally have: $$ \int_{0}^{+\infty}J_0(x)e^{-px}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}=\frac{1}{p}\sum_{n\geq 0}\binom{-1/2}{n}\frac{1}{p^{2n}} $$ or $$ \int_{0}^{+\infty}J_0(x)e^{-px}\,dx = \frac{1}{p\sqrt{1+\frac{1}{p^2}}}=\frac{1}{\sqrt{p^2+1}}$$

how are these equal: $$\sum_{n\geq 0}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}=\frac{1}{p}\sum_{n\geq 0}\binom{-1/2}{n}\frac{1}{p^{2n}} ?$$

Answer 1

$$\begin{array}{rcl} \dbinom{-1/2}n &=& \dfrac {\left( -\frac12 \right) \left( -\frac32 \right) \left( -\frac52 \right) \cdots \left( -\frac{2n-1}2 \right)} {n!} \\ &=& \left( -\dfrac12 \right)^n \dfrac {1 \times 3 \times 5 \times \cdots \times (2n-1)} {n!} \\ &=& \left( -\dfrac12 \right)^n \dfrac {1 \times 2 \times 3 \times \cdots \times (2n)} {(2 \times 4 \times 6 \times \cdots \times (2n)) n!} \\ &=& (-1)^n \left( \dfrac12 \right)^{2n} \dfrac {1 \times 2 \times 3 \times \cdots \times (2n)} {(1 \times 2 \times 3 \times \cdots \times n) n!} \\ &=& (-1)^n \left( \dfrac12 \right)^{2n} \dfrac {(2n)!} {n! n!} \\ &=& (-1)^n \left( \dfrac12 \right)^{2n} \dbinom {2n} n \\ \end{array}$$