I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further expand the binomial coefficient and tried to write it as sum of integrals but a pesky "r" term ends up being multiplied
How can we evaluate this with and without using calculus?
Best Answer
Note that $$\sum_{r=0}^{\infty} {2r \choose r} x^r=(1-4x)^{-1/2}~~~~(1)$$ Proof:
This is due to binomial expansion of $(1-4x)^{-1/2}$ in infinite series as $$(1-4x)^{-1/2}=\sum_{r=0}^{\infty} {-1/2 \choose r} x^r~~~~(2)$$ Use $(1+z)^{p}=1+pz+p(p-1)z^2/2+p(p-1)(p-2)z^3/3!+\dots~,~ |z|<1.$
Next use ${-q \choose k}=(-1)^k{n+q-1 \choose k}$
So $$(1-4x)^{-1/2}=\sum_{r=0}^{\infty} (-1)^r{r+1/2-1 \choose r}(-4x)^r$$ $$\implies (1-4x)^{-1/2}=\sum_{r=0}^{\infty} 2^{2r} {r-1/2 \choose r} x^r$$ $$\implies (1-4x)^{-1/2} =\sum_{r=0}^{\infty} \frac{1\cdot2\cdot3\cdot 4\cdots(2r-1)\cdot 2r}{r!~ r!}x^r=\sum_{r=0}^{\infty} {2r \choose r} x^r$$