Solutions to MATH 152 Fall 2008 Exam 3A
1.
B n→∞
lim (a2n − 3bn ) = ( lim an )2 − 3 lim bn = 22 − 3(−3) = 13.
n→∞
n→∞
ln x
lim
C Apply L'Hospital's Rule to x→∞
= lim x = 0
x→∞ 1
x
3. D Complete
the square: x2 + (y 2 − 2y + 1) + z 2 = 1 + 1; x2 + (y − 1)2 + z 2 = 2, so r2 = 2 and
√
1
2.
2.
r=
4.
C Using the Comparison Test.
essarily mean that
∞
X
(D) is NOT neessarily true beause lim an = 0 does not nen→∞
an is onvergent.
n=1
5.
n
n
= 1, the sequene (−1)n
B Sine n→∞
lim
alternates between −1 and 1, therefore the
n+1
n+1
terms of the series do not approah 0, whih means the series diverges by the Test for Divergene.
B
∞
∞
X
X
1
bn
bn and
an either both onverge
= 1, whih means the series
6. Let an = . Then lim
n→∞ an
n
n=1
n=1
∞
∞
X
X
or both diverge. Sine
an diverges (by Integral Test or P-Test),
bn diverges by the Limit
Comparison Test.
n=1
n=1
7.
A
8.
C Sine the terms of both series approah zero, both ∞series onverge by ∞the Alternating Series
∞ X
1 1
1
1
1 1
1 1
1
, whih is a Telesoping Series: sN =
+
+
=
−
−
−
−
(n + 2)(n + 3) n=1 n + 2 n + 3
3 4
4 5
5 6
n=1
as N → ∞.
∞
X
Test. To test absolute onvergene, we look at (IA)
X
n=1
1
n3/4
and (IIA)
X
n=1
1
n4/3
. Both are P -
series; in (IA), P < 1so the series diverges, and in (IIA), P > 1 so the series onverges. Therefore,
series (I) onverges but not absolutely, and series (II) onverges absolutely.
9.
B The series an be written as
∞ n−1
X
4
2
3
n=1
4
2
r = . The sum of the series is 3
3
1−
10.
C
2
3
The Malaurin series for cos x is
∞
∞
X
X
(−1)n x4n
(−1)n (x2 )2n
=
.
(2n)!
(2n)!
n=0
n=0
3
, whih is a geometri series with a =
4
and
3
= 4.
∞
X
(−1)n x2n
.
(2n)!
n=0
So the Malaurin series for cos(x2 ) is
11. (a) AB =< 1, 2, 2 >, BC =< 0, −1, 1 >. AB · BC = (1)(0) + (2)(−1) + (2)(1) = 0, so the sides
are perpendiular to eah other.
(b) |AB| =
√
3 2
.
2
p
√
√
1
12 + 22 + 22 = 3, |BC| = 02 + (−1)2 + 12 = 2, so the area is |AB||BC| =
2
1
12. f (−1) = 4; f ′ (x) = 8x3 − 1, so f ′ (−1) = −9; f ′′ (x) = 24x2 , so f ′′ (−1) = 24; f ′′′ (x) = 48x, so
f ′′′ (−1) = −48. The third degree Taylor Polynomial is f (−1) +
1)2 +
f ′′′ (−1)
(x + 1)3 = 4 − 9(x + 1) + 12(x + 1)2 − 8(x + 1)3
3!
f ′′ (−1)
f ′ (−1)
(x + 1) +
(x +
1!
2!
∞
∞
X
X
xn
(−1)n
, so e−1 =
. Therefore, subtrating the rst term of the series gives us
n!
n!
n=0
n=0
∞
X
(−1)n
e−1 − 1 =
.
n!
n=1
13. ex =
n+1
(x−1)
√
2n+1 √n+1 n
|x − 1|
14. (a) Applying the Ratio Test gives us absolute onvergene when lim (x−1)n = lim
·√
<1
n→∞ n→∞
2
n+1
√
2n n
and divergene when the limit is > 1. Sine the seond fration approahes 1, we have ab|x − 1|
< 1, |x − 1| < 2 whih makes the radius of onvergene
solute onvergene when
2
2.
(b) The series onverges when −2 < x − 1 < 2, −1 < x < 3. To nd the interval of onvergene,
∞
∞
X
X
(−1)n
(−2)n
√ , whih on√
=
n
n
2 n
n=1
n=1
∞
∞
X 2n
X
1
√ ,
√
verges by the Alternating Series test. Whne x = 3, the series beomes
=
n
2 n n=1 n
n=1
test the endpoints: When x = −1, the series beomes
whih diverges by the P-test or integral test. Therefore, the interval of onvergene is
−1 ≤ x < 3.
∞ ∞
X
X
(−1)n+1
(−1)n+1 2n
x dx = C +
x2n+1
(2n
+
1)!
(2n
+
1)!
(2n
+
1)(2n
+
1)!
n=1
n=1
n=1
2n+1
1/2
∞
∞
n+1
n+1
X
X
(−1)
1
(−1)
2n+1 1/2
x
|0 =
S(x) dx =
(b)
.
(2n
+
1)(2n
+
1)!
(2n
+
1)(2n
+
1)!
2
0
n=1
n=1
2·4+1
9
1
1
1
1
=
() Sine the series is alternating |S−S3 | ≤ |a4 | =
.
(2 · 4 + 1)(2 · 4 + 1)! 2
9 · 9! 2
15. (a)
S(x) dx =
X
∞
(−1)n+1
x2n dx =
2