MATH 172.504
Examination 3 Solutions
April 26, 2012
1.
Find the formula for the n-th term of the sequence {an } that starts
1
1
3
9
27
81
a0 = − , a1 = , a2 = − , a3 = , a4 = − , a5 = .
3
4
7
10
13
16
(A)
(B)
(C)
(D)
(E)
(−1)n · (3n − 3n−1 )
3n − 1
n
(−1) · 3n
n+3
(−1)n+1 · 3n−1
3n + 1
(−1)n+1 · 3n2
3n + 1
(−1)n+1 · 3n+1
3n + 9
Solution: C It is a simple matter to check that this is correct.
2.
3 cos(2πn)
.
1
n→∞
n −2
Evaluate lim
(A) 0
(B) 3
1
2
3
(D) −
2
(E) The limit does not exist.
(C) −
Solution: D The key thing is that, when n is an integer, 2n + 1 is an odd integer, and so
cos(2nπ) = 1. So the limit is then
3.
The series
3
−2
= − 23 .
∞
X
32n + (−2)n+1
10n
n=0
25
3
17
(B) converges to
2
(C) converges to 8
(A) converges to
(D) converges to 9
(E) diverges.
Solution: A This series can be decomposed into two geometric series:
n
∞
∞ ∞
X
32n + (−2)n+1 X 32 n X
−2
=
+
−2 ·
.
10n
10
10
n=0
n=0
1
n=0
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
of S. We therefore want the smallest m for which
is within 11m+1
1
1
< 0.001 = 3 .
11m+1
10
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 p
(A) 2π
x 1 + 16x2 dx
1
3
Z
(B) 4π
x2
p
1 + 16x2 dx
x2
p
1 − 16x2 dx
1
3
Z
(C) 4π
1
Z
3
(D) 2π
Z
√
x 1 − 4x dx
1
3
(E) 4π
x2
p
1 + 4x4 dx
1
Solution: B Using the formula for the surface area of a surface of revolution,
3
Z
2x2
S.A. = 2π
1
which is B.
2
p
1 + (4x)2 dx,
6.
Find the arclength of the graph of the curve defined by y = 2x3/2 from x = 0 to x = 5/3.
Solution: We use the arclength formula:
Z
5/3 p
1+
s=
(y 0 )2 dx
5/3 q
√
1 + (3 x)2 dx
Z
=
0
0
5/3 √
Z
=
1 + 9x dx
0
16 √
Z
=
1
=
=
7.
u du (u = 1 + 9x, du = 9 dx)
16
2 3/2
u
27
1
2
2
14
(16)3/2 −
=
.
27
27
3
Find the limits of the following sequences.
2n3 − 3
(a) an = √
9n6 + 20n
n−1 n
(b) bn =
n
Solution: (a) By comparing highest order terms in the numerator and denominator (both
are degree 3), we see that the limit will be
√2
9
=
2
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
!
ln 1 − n1
= lim exp
1
n→∞
= exp
n
lim
= exp  lim
n→∞
= exp(−1) =
1
n
!
1
n
n→∞

8.
ln 1 −
1
1
1− n
·
− n12
1
n2

(l’Hôpital’s Rule)

1
.
e
[10 points] Consider the series
∞ X
n=1
1
1
−
2n + 1 2n + 3
(a) Find a formula for the partial sum Sm of the series.
(b) Find the sum of the series or show that it diverges.
3
.
Solution: (a) This is a telescoping series. After canceling the intermediate terms we have
1
1
.
−
3 2m + 3
Sm =
(b) To find the sum of the series we take the limit of Sm as m → ∞:
∞ X
π
1
π
cos
− cos
= lim Sm =
.
m→∞
n
n+2
3
n=1
9.
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
Solution: (a) First we observe that cos(n)
≤
n2 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 .
converges. Therefore, by the absolute convergence test , the series
n2
(b) Because
1/n 0
1
1
lim
=
= 1 6= 0,
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,
3
n→∞ ((n + 1)!)
(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
(x + 1)n
√
.
n
n
·
3
n=1
(a) What is the open interval of convergence for f (x)?
(b) Does the series converge at the endpoints of the interval in (a)? Why or why not?
4
(c) Write down a power series expression for g(x) = f 00 (x).
Solution: (a) We apply the ratio test:
√
√
|x + 1|
n
|x + 1|n+1
3n n
|x + 1|
= lim
lim √
·
·√
=
.
n
n+1
n→∞
n→∞
|x + 1|
3
3
n+1·3
n+1
So the open interval of convergence is defined by
|x + 1|
< 1 ⇐⇒ −3 < x + 1 < 3 ⇐⇒ −4 < x < 2 .
3
(b) We check at the endpoints. When x = 2, the series we obtain is
∞
X
1
√ ,
n
n=1
which diverges as it is a p-series with p = 1/2 ≤ 1 . When x = −4, the series we obtain is
∞
X
(−1)n
√ ,
n
n=1
which converges by the alternating series test .
(c) We differentiate the series for f (x) term-by-term and find
f 00 (x) =
∞
X
n(n − 1)(x + 1)n−2
√
.
n
n
·
3
n=2
5