Homework 11.8: pp 769--, #14, 16,  38

#14:

(∑_(n=1)^∞)(((((-1))^n)(x^(2n+1)))/((2n+1)!))

a[n_]:=(((((-1))^n)(x^(2n+1)))/((2n+1)!))
r[n_]:=Simplify[Abs[a[n+1]/a[n]]]
r[n]
L=Limit[r[n],n->Infinity]

Since the limit of r is zero, the radius of convergence is infinite, and the interval of convergence is all real numbers.

#16

c[n_]:=((((-1))^n)/((2n+1)))
c[n]
s[x_]:=(∑_(n=1)^∞)c[n](((x-3))^n)

r[n_]:=Simplify[Abs[c[n+1]/c[n]]]
r[n]
L=Limit[r[n],n->Infinity]
s[4]

Since the limit of r is 1, the radius of convergence is 1, and the interval of convergence is (2,4]. The right-hand endpoint is included, because the series alternates, with decreasing positive terms (aside from the alternation).

#38

f[x_]:=(c0+c1*x+c2*x+c3*x)(1/((1-x^4)))

with radius of convergence 1, and interval of convergence (-1,1).