Question

If f^(n) (0) = (n+1)! for n = 0,1,2,..., find the Maclaurin series for f and its radius of convergence.

Answer 1

Notice that the Maclaurin series for f(x) is

f(x)=∞∑n=0f(n)(0)xnn!=∞∑n=0(n+1)!xnn!=∞∑n=0(n+1)n!xnn!=∞∑n=0(n+1)xnf(x)=∑n=0∞f(n)(0)xnn!=∑n=0∞(n+1)!xnn!=∑n=0∞(n+1)n!xnn!=∑n=0∞(n+1)xn

Because

limn→∞∣∣∣(n+2)xn+11⋅1(n+1)xn∣∣∣=|x|limn→∞n+2n+1=|x|⋅1=|x|limn→∞|(n+2)xn+11⋅1(n+1)xn|=|x|limn→∞n+2n+1=|x|⋅1=|x|

Convergence will occur provided that

|x|<1−1<x<1|x|<1−1<x<1

thereby giving us a radius of convergence as

R=1−(−1)2=1