Q:

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

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 $|\frac{a_n}{a_{n+1}}|$?

A:

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

