Calculus II
Exam 3, Fall 2002, Answers
1. Find the limits
ln x 1
a) lim x e ln ln x 1
x
x e 1 1
ln x x
Answer. l H lim
x 1 2x b) lim
x ∞ 3x2 1
Answer. lim lnx ln e 1 x e
x 2x2
∞ 3x2 1
lim
x
lim
x ∞
2
3
1
x
1
x2
2
3
2. Does the integral converge or diverge? Give reasons. If you can, evaluate the integral.
1
a)
0
dx
x9
10
Answer. ∞
b)
0
1
lim
a 0
a
x 9
10
dx lim 10x1 a 0
10 1
a
lim 10 1 a1
a 0
10
10 x
dx
1 x3
Answer. This converges because
x
1 x3
and
1
x2
∞
0
1
x
1
x2
dx
∞
x2 3. Does the series converge or diverge? Give reasons.
∞
a)
n 2n 1 ∑ 3n
n 0
Answer. This series converges by comparison with
∞
2
n 0
3
∑n
n
which is 2 3 times the derived geometric series. Here is the comparison:
2
2
n 2n 1 n
n n n n 3n
3
3
3
∞
b.
n
1
∑ ln n
n 0
Answer. This diverges by comparison with the series ∑ n 1 , since ln n
n.
4. What is the radius of convergence of the power series? Show your work.
∞
a)
∑
n 0
2n 1 xn
Answer. We calculate the ratios of the coefficients:
an 1
an
1
2n 1 1
2
n
2 1
1
1
2n 1
1
2n
2
1
R
so R 1 2.
∞
b)
3n
∑ n! xn
n 0
Answer. We calculate the ratios of the coefficients:
an 1
an
3n 1 n!
n 1 ! 3n
3
0
n 1 so R ∞.
5. Find the Maclaurin series for the function.
1 x
1 x
a)
Answer. We start with the geometric series:
∞
1
∑ xn
1 x
n 0
and then multiply by 1 x:
1 x
1 x
1
1 x
x
1 x
x
b)
0
∞
∞
∑ xn x ∑ xn n 0
∞
∞
n 0
n 1
n 0
∞
∑ xn ∑ xn n 1
Answer. We start with the geometric series:
∞
∑ xn
1 x
n 0
and substitute t 3 for x:
1
1 t3
∞
∑ t 3n
n 0
and finally, integrate (term by term on the right):
x
0
dt
1
t3
∞
n 0
n 0
1 2 ∑ xn
dt
1 t3
1
∞
∑ xn ∑ xn ∞
t 3n ∑ 3n n 0
1
1
1