<s>[INST] 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 $|\frac{a_n}{a_{n+1}}|$?
 [/INST] 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 $1$.
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