Start from the MacLaurin series for the exponential function:
#e^t = sum_(k=0)^oo t^k/(k!)#
and using the exponential formula for the hyperbolic cosine:
#cosh t = 1/2(e^t+e^(-t))#
#cosh t = 1/2(sum_(k=0)^oo t^k/(k!) + sum_(k=0)^oo (-t)^k/(k!) )#
#cosh t = 1/2sum_(k=0)^oo t^k/(k!) (1+ (-1)^k )#
We can see that for #k# even # (1+ (-1)^k ) = 2# while for #k# odd # (1+ (-1)^k ) = 0#.
So all the odd index terms are null, and if we let #k = 2n#:
#cosh t = sum_(n=0)^oo t^(2n)/((2n)!) #
Finally let #t=x^2#
#cosh (x^2) = sum_(n=0)^oo(x^2)^(2n)/((2n)!) = sum_(n=0)^oo x^(4n)/((2n)!) #
Using the ratio test we can see that:
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( (x^(4(n+1))/((2(n+1))!) )/ (x^(4n)/((2n)!) ))#
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( x^(4n+4)/x^(4n) ((2n)!)/((2n+2)!)#
#lim_(n->oo) abs(a_(n+1)/a_n) = x^4 lim_(n->oo) 1/((2n+2)(2n+1)) = 0#
whatever is the value of #x#, and we can conclude that the series has radius of convergence #R=oo#.
