Find the sum of the series $1+\frac12 z^2+\frac{1\cdot3}{2\cdot4}z^4+....$

Question

Find the sum of the series $1+\frac12 z^2+\frac{1\cdot3}{2\cdot4}z^4+....$. We can rewrite n-th term as $a_n=\frac{(n-1)(n-3)\cdot...\cdot1}{n(n-2)\cdot..\cdot2}z^n=\frac{(2n)!}{2^{2n}\cdot(n!)^2}z^n$. But I don't know what should I do now. Consider the ratio $|\frac{a_n}{a_{n+1}}|$?

Answer 1

Your series is$$\sum_{n=0}^\infty(-1)^n\binom{-\frac12}nz^{2n}$$and therefore its sum is $(1-z^2)^{-\frac12}$, when $|z|<1$.

Answer 2

Another approach (same origin as the other answer, but starting from a well-known formula): we have, for a real $z$ s. t. $|z|<1$, $$\arcsin z=z+\frac12\frac{z^3}3+\frac{1\cdot 3}{2\cdot 4}\frac{z^5}5+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{z^7}7+\dotsm, $$ whence $$(\arcsin z)'=\frac1{\sqrt{1-z^2\strut}}=1+\frac12z^2+\frac{1\cdot 3}{2\cdot 4}z^4+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}z^6+\dotsm. $$ The radius of convergence, of course, is $1$.

Answer 3

Due to De Moivre's formula $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$, the binomial theorem and the orthogonality relation $\int_{0}^{2\pi}e^{ni\theta}e^{-mi\theta}\,d\theta=2\pi\delta(m,n)$ we have $$ \int_{0}^{2\pi}\left(\cos\theta\right)^{2n}\,d\theta = 2\pi\frac{(2n)!}{4^n n!^2}\tag{1}$$ for any $n\in\mathbb{N}$, hence $$\sum_{n\geq 0}\binom{2n}{n}\frac{z^n}{4^n}=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{d\theta}{1-z\cos^2\theta}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-z\cos^2\theta}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{(1-z)+u^2}\tag{2}$$ and $\sum_{n\geq 0}\binom{2n}{n}\frac{z^n}{4^n}=\frac{1}{\sqrt{1-z}}$ holds for any $z\in(-1,1)$.
As an interesting consequence, by squaring both sides we get $$\forall n\in\mathbb{N},\qquad \sum_{m=0}^{n}\binom{2m}{m}\binom{2n-2m}{n-m}=4^n.\tag{3} $$ You may also go in the opposite direction: a combinatorial proof of $(3)$ leads to $\sum_{n\geq 0}\binom{2n}{n}\frac{z^n}{4^n}=\frac{1}{\sqrt{1-z}}$ almost immediately.