MATH 172.200
Examination 3 Solutions
April 26, 2012
1.
Find the formula for the n-th term of the sequence {an } that starts
3
5
7
a0 = 1, a1 = 1, a2 = − , a3 = , a4 = − .
2
6
24
(−1)n · (n + 1)!
2n
n
(−1) · (n + 2)
(B)
2n + 1
(−1)n · (n + 1)
(C)
n!
n+1
(−1)
· (2n + 1)
(D)
(n + 1)!
(A)
(−1)n+1 · (2n − 1)
n!
Solution: E It is a simple matter to check that this is correct.
(E)
2.
Evaluate lim
n→∞
(A) −
3 cos((2n + 1)π)
.
1
n −2
3
2
3
2
(C) −3
(B)
(D) 3
(E) The limit does not exist.
Solution: B The key thing is that, when n is an integer, 2n + 1 is an odd integer, and so
cos((2n + 1)π) = −1. So the limit is then
3.
The series
∞
X
32n + (−2)n+1
n=0
10n
(A) converges to 9
25
(B) converges to
2
25
(C) converges to
3
35
(D) converges to
3
(E) diverges.
1
−3
−2
=
3
2
.
Solution: C This series can be decomposed into two geometric series:
∞
X
32n + (−2)n+1
=
10n
n=0
∞ 2 n
X
3
n=0
10
+
∞
X
−2 ·
n=0
−2
10
n
.
9
Since both geometric series converge (| 10
| < 1 and | −2
10 | < 1), from the formula for the sum
of a geometric series,
∞
X
32n + (−2)n+1
n=0
4.
∞
X
=
10n
1
2
9 −
1 − 10
1+
1
5
=
25
.
3
1
. By the Alternating Series Estimation Theorem,
11n
n=1
what is the smallest number of terms m of this series that we would need to take so that
the partial sum
m
X
1
(−1)n+1 n
11
Consider the series S =
(−1)n+1
n=1
is within 0.001 of the value of the sum S?
(A) 1
(B) 2
(C) 3
(D) 4
(E) The series diverges.
Solution: B By the Alternating Series Estimation Theorem, the sum of the first m terms
1
is within 11m+1
of S. We therefore want the smallest m for which
1
11m+1
< 0.001 =
1
.
103
So we need m = 2 .
5.
Which of the following integrals represents the surface area of the surface obtained by
rotating the graph of y = 2x2 over the interval [1, 3] about the x-axis?
Z 3
√
(A) 2π
x 1 − 4x dx
1
3
Z
x2
(B) 4π
p
1 + 4x4 dx
1
3
Z
(C) 2π
x
p
1 + 16x2 dx
1
Z
3
(D) 4π
Z
x2
1
3
(E) 4π
x2
p
1 + 16x2 dx
p
1 − 16x2 dx
1
2
Solution: D Using the formula for the surface area of a surface of revolution,
3
Z
2x2
S.A. = 2π
p
1 + (4x)2 dx,
1
which is D.
6.
Find the arclength of the graph of the curve defined by
y=
x4
1
+ 2
16 2x
from x = 1 to x = 2.
Solution: We use the arclength formula:
s
3
2
Z 2p
Z 2
x
1
s=
1 + (y 0 )2 dx =
1+
− 3 dx
4
x
1
1
Z 2r
x6 1
1
1+
− + 6 dx
=
16 2 x
1
Z 2r 6
x
1
1
+ + 6 dx
=
16 2 x
1
s
2
Z 2 3
x
1
=
+ 3 dx
4
x
1
Z 2 3
1
x
+ 3 dx
=
4
x
1
4
2 x
1
1
1
1
21
=
− 2 = 1−
−
−
=
.
16 2x 1
8
16 2
16
7.
[10 points] Find the limits of the following sequences.
p
p
(a) an = n2 + 10n + 1 − n2 + 5n + 1
n−1 n
(b) bn =
n
Solution: (a)
p
p
lim ( n2 + 10n + 1 − n2 + 5n + 1)
n→∞
√
√
p
p
n2 + 10n + 1 + n2 + 5n + 1
2
2
√
= lim ( n + 10n + 1 − n + 5n + 1) · √
n→∞
n2 + 10n + 1 + n2 + 5n + 1
(n2 + 10n + 1) − (n2 + 5n + 1)
√
= lim √
n→∞
n2 + 10n + 1 + n2 + 5n + 1
5n
5
√
= lim √
=
.
2
2
n→∞
2
n + 10n + 1 + n + 5n + 1
3
(b) There are several ways to do this problem, but one is the following:
n−1 n
1
lim
= lim exp n ln 1 −
n→∞
n→∞
n
n
!
1
ln 1 − n
= lim exp
1
n→∞
= exp
n
lim
= exp  lim
n→∞
= exp(−1) =
Consider the series
1
n
!
1
n
n→∞

8.
ln 1 −
1
1
1− n
·
− n12
1
n2


(l’Hôpital’s Rule)
1
.
e
∞ X
π
π
cos
− cos
.
n
n+2
n=1
(a) Find a formula for the partial sum Sm of the series.
(b) Find the sum of the series or show that it diverges.
Solution: (a) This is a telescoping series. After canceling the intermediate terms we have
π π
π
Sm = cos(π) + cos
− cos
− cos
2
m+1
m+2
π
π
− cos
= −1 − cos
.
m+1
m+2
(b) To find the sum of the series we take the limit of Sm as m → ∞:
∞ X
n=1
9.
π
π
− cos
= lim Sm = −1 − 1 − 1 = −3 .
cos
m→∞
n
n+2
Determine whether each series below converges or diverges. For each problem use one or
more of the tests we have studied: absolute convergence test, alternating series test, direct
comparison test, geometric series test, integral test, limit comparison test, p-series test,
ratio test, or test for divergence (n-th term test). Be sure to specify which test(s) you are
using and show how they apply.
∞
X
cos(n)
(a)
n2
n=1
∞ 1/n
X
1
(b)
2
(c)
n=1
∞
X
n=1
(3n)!
(n!)3
4
Solution: (a) First we observe that cos(n)
≤
2
n
1
n2
for all n, and since the series
1
conn2
P cos(n) n2 P
verges (it is a p-series with p = 2 > 1), by the direct comparison test , the series
P cos(n)
converges. Therefore, by the absolute convergence test , the series
converges .
n2
(b) Because
1/n 0
1
1
=
= 1 6= 0,
lim
n→∞ 2
2
the test for divergence (n-th term test) implies that the series must diverge .
(c) We apply the ratio test :
(3(n + 1))! (n!)3
(3n + 3)(3n + 2)(3n + 1)
·
= lim
= 27 > 1,
n→∞ ((n + 1)!)3 (3n)!
n→∞
(n + 1)3
ρ = lim
and so the ratio test implies that the series diverges .
10.
Consider the following function:
f (x) =
∞
X
(2x + 1)n
n=0
3n+1
.
(a) What is the open interval of convergence of f (x)?
Z
(b) Write down a power series expression for f (x) dx centered about the midpoint of the
interval in (a).
(c) Does the series in (b) converge at the endpoints of the interval in (a)?
ax + b
for all x in the interval in (a).
(d) Find constants a, b, c, d so that f (x) =
cx + d
Solution: (a) We apply the ratio test:
n+1
(2x + 1)n+1
3
= lim |2x + 1| = |2x + 1| .
ρ = lim ·
n+2
n
n→∞
3
(2x + 1) n→∞
3
3
So the open interval of convergence is defined by
|2x + 1|
< 1 ⇐⇒ −3 < 2x + 1 < 3
3
⇐⇒ −4 < 2x < 2
⇐⇒ −2 < x < 1 .
(b) We integrate the series for f (x) term-by-term:
Z
f (x) dx = C +
∞
∞
X
X
2n (x + 12 )n+1
2n−1 (x + 21 )n
=
C
+
.
(n + 1)3n+1
n · 3n
n=0
n=1
(c) We check at the endpoints. When x = −2, the series we obtain is
∞
X
2n−1 · (− 3 )n
2
n=1
n · 3n
5
=
∞
X
(−1)n
n=1
2n
.
This series converges , as it is a constant multiple of the alternating harmonic series .
When x = 1, the series we obtain is
∞
X
2n−1 · ( 3 )n
n=1
n·
2
3n
∞
X
1
=
.
2n
n=1
This series diverges , as it is a constant multiple of the harmonic series .
(d) Here we use the fact that f (x) is a geometric series:
f (x) =
∞
X
1 2x + 1 n
n=0
3
3
6
=
1
1
3
− 2x+1
3
=
1
.
2 − 2x