Start from the binomial series:
#1/sqrt(1-t) = (1-t)^(-1/2) = sum_(n=0)^oo -1/2(-1/2-1)...(-1/2-n+1)(-t)^n/(n!)#
simplifying:
#1/sqrt(1-t) = (1-t)^(-1/2) = sum_(n=0)^oo -1/2(-3/2)...(-(2n-1)/2)(-t)^n/(n!)#
#1/sqrt(1-t) = sum_(n=0)^oo (-1)^n (1*3* (2n-1))/(2^n(n!))(-t)^n#
#1/sqrt(1-t) = sum_(n=0)^oo ((2n-1)!!)/(2^n(n!))t^n#
Applying the ratio test we see that:
#lim_(n->oo) abs(a_(n+1)/a_n ) = lim_(n->oo)abs ( (((2n+1)!!)/(2^(n+1)((n+1)!))t^(n+1))/( ((2n-1)!!)/(2^n(n!))t^n))#
#lim_(n->oo) abs(a_(n+1)/a_n ) = lim_(n->oo) ((2n+1)/(2n+2)) abs t = abs(t)#
so the series has radius of convergence #R=1#.
Let now #t = u^2# and as #abs t < 1 => x^2 < 1# we have:
#1/sqrt(1-u^2) = sum_(n=0)^oo ((2n+1)!!)/(2^n(n!))u^(2n)#
still with #R=1#.
Inside the interval #x in (-1,1)# we can then integrate term by term and obtain a series with the same radius of convergence:
#int_0^v (du)/sqrt(1-u^2) = sum_(n=0)^oo ((2n+1)!!)/(2^n(n!))int_0^v u^(2n)du#
#arcsinv= sum_(n=0)^oo ((2n-1)!!)/(2^n(n!)) v^(2n+1)/(2n+1)#
Note that:
# (2n-1)!! = ((2n-1)(2n-3)...3*1) = ((2n)(2n-1)(2n-2)(2n-3)...3*2*1)/((2n)(2n-2)...2)#
# (2n-1)!! = ((2n)!)/(2^n(n(n-1)...1)) = ((2n)!)/(2^n(n!))#
so we can write the series also as:
#arcsinv= sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) v^(2n+1)/(2n+1)#
Finally let: #v=x^3#. Again as #abs v<1 => abs(x^3) < 1# the radius of convergence does not change:
#arcsin(x^3) = sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) (x^3)^(2n+1)/(2n+1)#
#arcsin(x^3) = sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) x^(6n+3)/(2n+1)#
