Millersville University Name Answer Key
Department of Mathematics
MATH 211, Homework 09
March 31, 2006
Page 681, Exercise 22
Determine the radius and interval of convergence of the series
โ
X
k=0
3k
k!xk.
Applying the Ratio Test for absolute convergence we find that
lim
kโโ
๎
๎
๎
๎
๎
๎
3k+1
(k+1)! xk+1
3k
k!xk
๎
๎
๎
๎
๎
๎
= lim
kโโ
๎
๎
๎
๎
3k+1
3k
k!
(k+ 1)!
xk+1
xk๎
๎
๎
๎
= lim
kโโ
3
k+ 1|x|
= 0 for all x.
Thus the radius of convergence is r=โand the interval of convergence is โโ <x<โ.
Page 681, Exercise 36
Determine the radius and interval of convergence of the series
โ
X
k=0
k2
k!(x+ 1)k.
Applying the Ratio Test for absolute convergence we find that
lim
kโโ
๎
๎
๎
๎
๎
๎
(k+1)2
(k+1)! (x+ 1)k+1
k2
k!(x+ 1)k
๎
๎
๎
๎
๎
๎
= lim
kโโ
๎
๎
๎
๎
(k+ 1)2
k2
k!
(k+ 1)!
(x+ 1)k+1
(x+ 1)k๎
๎
๎
๎
= lim
kโโ
k+ 1
k2|x+ 1|
= 0 for all x.
Thus the radius of convergence is r=โand the interval of convergence is โโ <x<โ.