(1)
x+3x3+5x5+⋯+2n+1x2n+1+⋯
∑n=0∞2n+1x2n+1
r=limn→∞2n+312n+11=limn→∞2n+12n+3=1
(2)
1+2!x2+4!x4+⋯+(2n)!x2n
∑n=0∞(2n)!x2n
r=limn→∞{2(n+1)}!1(2n)!1=limn→∞(2n)!{2(n+1)}!=limn→∞∣(2n+1)(2n+2)∣=∞
(3)
x+32!x2+53!x3+⋯+2n−1n!xn+⋯
∑n=1∞2n−1n!xn
r=limn→∞2n+1(n+1)!2n−1n!=limn→∞(2n−1)(n+1)!(2n+1)n!=limn→∞(2n−1)(n+1)2n+1=limn→∞2n2+n−12n+1=0