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Quiz 4: Convergence and Summability of Sequences and Series in Math 253, Quizzes of Advanced Calculus

A math quiz focused on the concepts of convergence and summability of sequences and series. Students are required to decide if certain statements about sequences are always, sometimes, or never true, and to determine if specific series converge or diverge. The quiz covers topics such as the comparison test and limit comparison test.

Typology: Quizzes

Pre 2010

Uploaded on 09/17/2009

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Quiz 4 (Practice) Name
Math 253
July 21, 2009
Show all work. Answers without adequate justification will not receive credit.
For problems 1-7, assume
โˆž
X
n=1
an= 5.
Decide if the following statements are ALWAYS true, SOMETIMES true, or NEVER true. Circle
the appropriate answer.
1.
๎˜ƒ
๎˜‚
๎˜€
๎˜
ALWAYS SOMETIMES NEVER The sequence {an}converges.
The sequence must converge to 0 since the series converges.
2.
๎˜ƒ
๎˜‚
๎˜€
๎˜
ALWAYS SOMETIMES NEVER The sequence {an}is summable.
Saying a sequence is summable is the same as saying the infinite series converges.
3. ALWAYS
๎˜ƒ
๎˜‚
๎˜€
๎˜
SOMETIMES NEVER The sequence {an}is absolutely summable.
โˆž
X
n=1
5(โˆ’1)n+1
nln(2) =5
ln(2)
โˆž
X
n=1
(โˆ’1)n+1
n= 5, but โˆž
X
n=1 ๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
5(โˆ’1)n+1
nln(2) ๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=โˆž
4. ALWAYS SOMETIMES
๎˜ƒ
๎˜‚
๎˜€
๎˜
NEVER The sequence {an}diverges.
5. ALWAYS SOMETIMES
๎˜ƒ
๎˜‚
๎˜€
๎˜
NEVER The sequence Sk=Pk
n=1 anis summable.
The limit of the sequence {Sk}is 5, so the series Pโˆž
k=1 Skcannot converge.
6. ALWAYS
๎˜ƒ
๎˜‚
๎˜€
๎˜
SOMETIMES NEVER The sequence {(โˆ’1)nan}is summable.
e.g. an= (โˆ’1)n1
n.
7.
๎˜ƒ
๎˜‚
๎˜€
๎˜
ALWAYS SOMETIMES NEVER The sequence {nโˆ’2an}is summable.
Since anโ†’0 as nโ†’ โˆž, the sequence anis bounded. In particular, there is some number M > 0
such that |an|< M for all n. Now we have
๎˜Œ
๎˜Œ
๎˜Œ
an
n2๎˜Œ
๎˜Œ
๎˜Œ=|an|
n2โ‰คM
n2
Since {M
n2}is summable, the sequence {๎˜Œ
๎˜Œnโˆ’2an๎˜Œ
๎˜Œ}is summable by the comparison test. In other
words {nโˆ’2an}is absolutely summable, and thus it is also summable.
pf2

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Quiz 4 (Practice) Name

Math 253

July 21, 2009

Show all work. Answers without adequate justification will not receive credit.

For problems 1-7, assume โˆž โˆ‘

n=

an = 5.

Decide if the following statements are ALWAYS true, SOMETIMES true, or NEVER true. Circle

the appropriate answer.

ALWAYS SOMETIMES NEVER The sequence {an} converges.

The sequence must converge to 0 since the series converges.

ALWAYS SOMETIMES NEVER The sequence {an} is summable.

Saying a sequence is summable is the same as saying the infinite series converges.

3. ALWAYS

SOMETIMES NEVER The sequence {an} is absolutely summable.

โˆž โˆ‘

n=

n+

n ln(2)

ln(2)

โˆž โˆ‘

n=

n+

n

= 5, but

โˆž โˆ‘

n=

n+

n ln(2)

4. ALWAYS SOMETIMES

NEVER The sequence {an} diverges.

5. ALWAYS SOMETIMES

NEVER The sequence Sk =

โˆ‘k

n= an is summable.

The limit of the sequence {Sk} is 5, so the series

k= Sk cannot converge.

6. ALWAYS

SOMETIMES NEVER The sequence {(โˆ’1) n an} is summable.

e.g. an = (โˆ’1)

n 1 n

ALWAYS SOMETIMES NEVER The sequence {n โˆ’ 2 an} is summable.

Since an โ†’ 0 as n โ†’ โˆž, the sequence an is bounded. In particular, there is some number M > 0

such that |an| < M for all n. Now we have

an

n 2

โˆฃ =^

|an|

n 2

M

n 2

Since {

M n^2 } is summable, the sequence {

n

โˆ’ 2 an

} is summable by the comparison test. In other

words {n

โˆ’ 2 an} is absolutely summable, and thus it is also summable.

For each of the following series, decide whether it is convergent or divergent.

โˆž โˆ‘

n=

cos(n)

n

n

Solution: Since โˆ’ 1 โ‰ค cos(n) โ‰ค 1 we have

cos(n)

n

n

n

n

The series โˆž โˆ‘

n=

n

n

is a p-series with p =

3 2

1 so it converges. By the comparison test {

cos(n) n

โˆš n

} is absolutely

summable and thus summable.

โˆž โˆ‘

n=

n + 1

6 n โˆ’ 5

Solution: Since lim nโ†’โˆž

n + 1

6 n โˆ’ 5

= 0, the series diverges.

โˆž โˆ‘

n=

n

2

  • 1

n 3

  • 1

Solution: We will use the limit comparison test with

an =

n 2

  • 1

n 3

  • 1

, bn =

n

We find the limit

lim nโ†’โˆž

an

bn

= lim nโ†’โˆž

n

3

  • n

n 3

  • 1

Since the limit exists and is greater than 0, and

n

is not summable,

n

2

  • 1

n^3 + 1

is not

summable either.

โˆž โˆ‘

n=

n

2

n!

Solution: We will use the ratio test. Let an = n

2 /n!, and find the limit

lim nโ†’โˆž

an+

an

= lim nโ†’โˆž

(n + 1)

2 n!

n 2 (n + 1)!

= lim nโ†’โˆž

n + 1

n

n + 1

2 0 = 0.

Since the limit is less than 1, the series converges absolutely.