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Solutions for Quiz 3 - Calculus II | MATH 1592, Quizzes of Calculus

Material Type: Quiz; Professor: Liu; Class: CALCULUS II; Subject: Mathematics; University: University of Central Arkansas; Term: Unknown 1989;

Typology: Quizzes

Pre 2010

Uploaded on 08/17/2009

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Math 1592 Solutions of Quiz 3
Problem 1. (10 points, 5 each) Determine the convergence of the following sequences.
If convergent, find the limits:
1. an=5n2
n2+2 .
Method 1: Use the L’Hˆopital’s Rule.
lim
n→∞
5n2
n2+ 2 = lim
x→∞
5x2
x2+ 2 = lim
x→∞
10x
2x= 5.
Method 2: Use limit properties.
lim
n→∞
5n2
n2+ 2 = lim
n→∞
5
1 + 2
n2
=5
1 + lim
n→∞
2
n2
= 5.
2. an= (1)n.
Diverges because the sequence alternates between -1 and 1.
Problem 2. (10 points, 5 each) Determine the convergence of the following series.
If convergent, find their sums:
1.
X
n=0
2µ3
4n
.
This is a geometric series. Because 3/4<1, it is convergent and its sum is
X
n=0
2µ3
4n
=2
13
4
= 8
2.
X
n=1
n2
n2+ 1
Since
lim
n→∞
n2
n2+ 1 = lim
n→∞
1
1 + 1
n2
=1
1 + lim
n→∞
1
n2
= 1 6= 0,
the series diverges.
1

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Math 1592 Solutions of Quiz 3

Problem 1. (10 points, 5 each) Determine the convergence of the following sequences. If convergent, find the limits:

  1. an = (^) n^52 n+2^2. Method 1: Use the L’Hˆopital’s Rule.

nlim→∞

5 n^2 n^2 + 2 = (^) xlim→∞

5 x^2 x^2 + 2 = lim x→∞

10 x 2 x

Method 2: Use limit properties.

nlim→∞

5 n^2 n^2 + 2 = (^) nlim→∞

1 + (^) n^22

1 + lim n→∞

n^2

  1. an = (−1)n. Diverges because the sequence alternates between -1 and 1.

Problem 2. (10 points, 5 each) Determine the convergence of the following series. If convergent, find their sums:

∑^ ∞ n=

)n .

This is a geometric series. Because 3/ 4 < 1, it is convergent and its sum is ∑^ ∞ n=

)n

∑^ ∞ n=

n^2 n^2 + 1 Since

nlim→∞

n^2 n^2 + 1 = (^) nlim→∞

1 + (^) n^12

1 + limn→∞

n^2

the series diverges.