Solutions to Quiz 2: Convergence of Series - Prof. Naoki Saito, Quizzes of Calculus

The solutions to quiz 2 questions related to determining the convergence or divergence of series using tests such as the direct comparison test and the root test.

Typology: Quizzes

Pre 2010

Uploaded on 12/09/2010

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Quiz 2 Solutions
Problem 1 (5 points): Determine if the series
1
X
n=0
sin2(n)
4nconverges or di-
verges. Give reasons for your answer.
Since 0sin2(n)
4n1
4n, and
1
X
n=0
1
4nconverges, being a geometric series, it
follows from the direct comparison test that
1
X
n=0
sin2(n)
4nconverges.
Problem 2 (5 points): Determine if the series
1
X
n=1
7(ln n)n
nnconverges or di-
verges. Give reasons for your answer.
By the root test,
1
X
n=1
7(ln n)n
nnconverges if limn!1 7(lnn)n
nn1
n= limn!1 71
nln n
n<
1. But limn!1 71
n= 70= 1 and limn!1 ln n
n= limn!1
1
n
1= 0, so that
limn!1 71
nln n
n= 0 <1. Therefore,
1
X
n=1
7(ln n)n
nnconverges.
1

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Quiz 2 Solutions

Problem 1 (5 points): Determine if the series

X^1

n=

sin^2 (n) 4 n^

converges or di-

verges. Give reasons for your answer.

Since 0 

sin^2 (n) 4 n^

4 n^

, and

X^1

n=

4 n^

converges, being a geometric series, it

follows from the direct comparison test that

X^1

n=

sin^2 (n) 4 n^

converges.

Problem 2 (5 points): Determine if the series

X^1

n=

7(ln n)n nn^

converges or di-

verges. Give reasons for your answer.

By the root test,

X^1

n=

7(ln n)n nn^

converges if limn!

7(ln n)n nn

n

= limn!1 7

1 n ln^ n n

1. But limn!1 7

1

n = 7^0 = 1 and limn!1 ln^ n

n

= limn!

1 n 1

= 0, so that

limn!1 7

1 n ln^ n n

= 0 < 1. Therefore,

X^1

n=

7(ln n)n nn^

converges.