MATH 2260 - 41514 MIDTERM EXAM 3 MODEL SOLUTION
SPRING 2013 - MOON
(1) (6 pts) Evaluate the series
∞
X
3
.
4n
n=2
n X
n
∞
∞
∞
X
X
3
1
1
3
=
3
3
=
−3−
n
4
4
4
4
n=2
n=2
n=0
=
3
1−
1
4
−3−
3
3
1
=4−3− =
4
4
4
1
• Describing the series as a geometric series with a = 3 and r = : 2 pts.
4
• Using a geometric series runs from n = 0 and first two terms, writing it as
n
∞
X
1
3
3
− 3 − : 4 pts.
4
4
n=0
1
• Evaluating the sum : 6 pts.
4
Date: April 11, 2013.
1
MATH 2260 - 41514 Midterm Exam 3
Spring 2013 - Moon
(2) (8 pts) Evaluate the limit
1
lim (en + n) n .
n→∞
1
1
1
(en ) n ≤ (en + n) n ≤ (2en ) n
1
lim (en ) n = lim e = e
n→∞
n→∞
1
n
1
1
n
1
lim (2en ) = lim 2 (en ) n = lim 2 n e = e
n→∞
n→∞
n→∞
1
⇒ lim (en + n) n = e
n→∞
1
1
• Making two appropriate sequences (en ) n and (2en ) n which give an upper
and lower bound of given sequence: 3 pts.
1
• Evaluating the limit of one side lim (en ) n = e: +2 pts.
n→∞
1
• Evaluating the limit of the other side lim (2en ) n = e: + 2 pts.
n→∞
1
• By using the sandwich theorem, obtaining the conclusion lim (en + n) n =
n→∞
e: 8 pts.
(3) (7 pts) Determine the convergence or divergence of the series
∞
X
n2 + 3
.
n3 + 4n
n=1
n2 + 3
n2
n2
1
∼
∼
= ,
3
3
3
n + 4n
n + 4n
n
n
∞
∞
X
X
1
n2 + 3
=∞⇒
should diverge
3 + 4n
n
n
n=1
n=1
n2 + 3
n2
n2
1
≥
≥
=
3
3
3
n + 4n
n + 4n
2n
2n
∞
∞
∞
X
X
1
1X1
n2 + 3
=
=∞⇒
=∞
3 + 4n
2n
2
n
n
n=1
n=1
n=1
n≥2⇒
• With appropriate reason, guessing the divergence : 3 pts.
n2 + 3
1
• Showing an inequality 3
≥
: 5 pts.
n + 4n
2n
∞
X
n2 + 3
• Proving
= ∞: 7 pts.
3 + 4n
n
n=1
2
MATH 2260 - 41514 Midterm Exam 3
Spring 2013 - Moon
(4) (9 pts) Determine the convergence or divergence of the series
∞
X
1
.
n
ln
n
n=2
1
1
is a positive decreasing function and f (n) =
x ln x
n ln n
Z ∞
∞
X
1
1
<∞⇔
dx < ∞
n
ln
n
x
ln
x
2
n=2
Z ∞
Z b
1
1
dx = lim
dx
b→∞ 2 x ln x
x ln x
2
1
u = ln x ⇒ du = dx, u(2) = ln 2, u(b) = ln b
x
Z b
Z ln b
1
1
b
dx =
du = [ln |u|]ln
ln 2 = ln | ln b| − ln | ln 2|
2 x ln x
ln 2 u
Z b
1
lim
dx = lim ln | ln b| − ln | ln 2| = ∞
b→∞ 2 x ln x
b→∞
Z ∞
∞
X
1
1
dx = ∞ ⇒
=∞
x ln x
n ln n
2
n=2
Z ∞
∞
X
1
1
Writing the precise statement of integral test
<∞⇔
dx <
n ln n
x ln x
2
n=2
∞ correctly: 2 pts.
Z b
1
Showing the definition of an improper integral lim
dx: 3 pts.
b→∞ 2 x ln x
Z b
1
By using substitution, evaluating the integral
dx = ln | ln b| −
2 x ln x
ln | ln 2|: 5 pts.
Evaluating the limit limb→∞ ln | ln b| − ln | ln 2| = ∞: 7 pts.
∞
X
1
Concluding
= ∞: 9 pts.
n ln n
n=2
f (x) =
•
•
•
•
•
3
MATH 2260 - 41514 Midterm Exam 3
(5) Let
Spring 2013 - Moon
∞
X
xn
f (x) =
.
n3n
n=1
(a) (6 pts) Find the radius of convergence.
∞ n ∞
X
x X
|x|n
=
n3n n3n
n=1
|x|n+1
(n+1)3n+1
|x|n
n→∞
n3n
lim
n=1
|x|n+1 n3n
|x|n
|x|
|x|n |x|n3n
= lim
=
=
lim
n
n+1
n
n
n→∞ |x| (n + 1)3
n→∞ 3(n + 1)
n→∞ |x| (n + 1)3 3
3
= lim
|x|
< 1 ⇔ |x| < 3 ⇒ radius of convergence = 3
3
• Taking the absolute value correctly and obtaining
∞
X
|x|n
n=1
• For the ratio test, setting up the limit
|x|n+1
(n+1)3n+1
lim
|x|n
n→∞
n3n
n3n
: 1 pt.
: 3 pts.
|x|
• Evaluating the limit and getting
: 5 pts.
3
• From the limit, obtaining the radius of convergence 3: 6 pts.
(b) (4 pts) Determine the convergence at endpoints.
∞
∞
∞
X
X
X
xn
3n
1
x=3⇒
=
=
=∞
n
n
n3
n3
n
n=1
n=1
n=1
∞
∞
X
X
xn
(−3)n
x = −3 ⇒
=
n3n
n3n
n=1
n=1
=
∞
X
(−1)n
n=1
n
converges by alternating series test
∞
X
1
• Finding
and stating the divergence of it: +2 pts.
n
n=1
∞
X
(−1)n
• Finding
and stating the convergence of it with appropriate
n
n=1
reason: +2 pts.
(c) (2 pts) Write down the interval of convergence.
[3, −3)
4
MATH 2260 - 41514 Midterm Exam 3
(6) (a) (7 pts) For f (x) =
√
3
Spring 2013 - Moon
x, find Taylor polynomial of f (x) of order 2 at x = 1.
√
3
f (1) = 1 = 1
1
1 2
f 0 (x) = x− 3 ⇒ f 0 (1) =
3
3
5
2
2 5
2
1
x− 3 = − x− 3 ⇒ f 00 (1) = −
f 00 (x) = · −
3
3
9
9
f 00 (1)
P2 (x) = f (1) + f (1)(x − 1) +
(x − 1)2
2!
0
− 92
1
1
1
= 1 + (x − 1) +
(x − 1)2 = 1 + (x − 1) − (x − 1)2
3
2!
3
9
• Finding f (1) = 1: +1 pt.
1
• Getting f 0 (1) = : +1 pt.
3
2
00
• Obtaining f (1) = − : +1 pt.
9
• Stating the definition of order 2 Taylor polynomial at x = 1 P2 (x) =
f 00 (1)
f (1) + f 0 (1)(x − 1) +
(x − 1)2 : + 2 pts.
2!
1
1
• Evaluating the answer 1 + (x − 1) − (x − 1)2 : 7 pts.
3
9
√
(b) (3 pts) By using (a), find an approximation of 3 1.1.
√
3
1.1 = f (1.1)
1
1
1
1
P2 (1.1) = 1 + (1.1 − 1) − (1.1 − 1)2 = 1 + (0.1) − (0.1)2
3
9
3
9
5
MATH 2260 - 41514 Midterm Exam 3
Spring 2013 - Moon
(7) (8 pts) The following table is a list of Taylor series of several functions.
function
Taylor series
∞
X xn
x2 x3
ex
=1+x+
+
+ ···
n!
2!
3!
n=0
∞
X
xn+1
x2 x3
ln(1 + x)
(−1)n
=x−
+
− ···
n
+
1
2
3
n=0
∞
X
x2n+1
x3 x5
sin x
(−1)n
=x−
+
− ···
(2n
+
1)!
3!
5!
n=0
∞
X
x2n
x2 x4
cos x
(−1)n
=1−
+
− ···
(2n)!
2!
4!
n=0
∞
X
1
xn = 1 + x + x2 + x3 + · · ·
1−x
n=0
Compute Taylor polynomial of
f (x) = e2x sin x
of order 3 at x = 0.
ex = 1 + x +
⇒ e2x = 1 + (2x) +
x2 x3
+
+ ···
2!
3!
(2x)2 (2x)3
4x2 8x3
+
+ · · · = 1 + 2x +
+
+ ···
2!
3!
2!
3!
x3 x 5
+
− ···
3!
5!
4x2 8x3
x3 x5
2x
f (x) = e sin x = 1 + 2x +
+
+ ···
x−
+
− ···
2!
3!
3!
5!
4x2 8x3
x3
4x2 8x3
= x 1 + 2x +
+
+ ··· −
1 + 2x +
+
+ ··· + ···
2!
3!
3!
2!
3!
4x3 8x4
x3 2x4
4x5
8x6
= x + 2x2 +
+
+ ··· −
−
−
−
− ··· + ···
2!
3!
3! 2!3! 3!3!
3!
4
1
= x + 2x2 +
−
x3 + · · ·
2! 3!
4
1
11
2
P3 (x) = x + 2x +
−
x3 = x + 2x2 + x3
2! 3!
6
sin x = x −
4x2 8x3
+
+ · · ·: 2 pts.
2!
3!
• By multiplyingpower series,
evaluating the Taylor series of f (x) and get
4
1
ting x + 2x2 +
−
x3 + · · ·: 6 pts.
2! 3!
11
• Computing the Taylor polynomial P3 (x) = x + 2x2 + x3 : 8 pts.
6
• Finding the Taylor series of e2x 1 + 2x +
6