We use the ratio test
If the #lim_(n->oo)∣a_(n+1)/a_n∣=L#
If #L<1#, then #sum_1 ^oa_n# converges
If #L>1#, then #sum_1 ^oa_n# diverges
If #L=1#, the test is inconclusive
#(∣a_(n+1)/a_n∣) =(∣(x^(2(n+1)+1)/((2(n+1)+1)!))/(x^(2n+1)/((2n+1)!))∣)#
Now, we calculate the limits
#lim_(n->oo)(∣(x^(2(n+1)+1)/((2(n+1)+1)!))/(x^(2n+1)/((2n+1)!))∣)#
#=lim_(n->oo)(∣(x^(2n+3)/((2n+3)!))/(x^(2n+1)/((2n+1)!))∣)#
#=lim_(n->oo)(∣(x^(2n+3)/(x^(2n+1))*(((2n+1)!)/((2n+3)!))∣)#
#=lim_(n->oo)(∣x^2*(1/((2n+3)(2n+2))∣)#
#=lim_(n->oo)(∣x^2*(1/((2n+3)2(n+1))∣)#
#=∣(x^2/2)∣*lim_(n->oo)(∣1/((2n+3)(n+1))∣)#
#=∣(x^2/2)∣*0#
#=0#
Therefore,
#L<1#, for every x,
So #sum_1 ^oox^(2n+1)/((2n+1)!)# converges for all x
The interval is # -oo < x <+oo #
