Ch 9

Question 1

{n} =

Answer 1

1, 2, 3, 4, 5, …

Question 2

{n²} =

Answer 2

1, 4, 9, 16, 25, 36, …

Question 3

{7.5 + 2.5(-1)ⁿ} =

Answer 3

5, 10, 5, 10, 5, 10, …

Question 4

1/3, 1/9, 1/27, 1/81, …

Answer 4

{1/3ⁿ}

Question 5

2, 3, 5, 7, 11, 13, 17, 19, 23, …

Answer 5

sequence of prime numbers, no explicit formula

Question 6

1, 1, 2, 3, 5, 8, 13, 21, …

Answer 6

F_n = F_n-1 + F_n-2

(Fibonacci Sequence)

Question 7

difference between a sequence and a series?

Answer 7

a sequence is a list of terms, while a series is the product of adding all terms together

Question 8

limit definition of a sequence

Answer 8

If a_n = f(n) for all positive integers, then:

the limit of f(x) as x approaches infinity = L

implies that

the limit of a_n as n approaches infinity = L

Question 9

what is the difference between a_n = n² and f(x) = x²?

Answer 9

they are the same except for domain, f(x) = x is continuous while a_n = n² is discreet a discreet set of points and so not continuous

Question 10

find the limit of a_n = (2n+5)^1/n

Answer 10

a_n is not continuous, so no derivitive can be taken. a_n must be rewritten as a function of x:

f(x) = (2x+5)^1/x yields indeterminate form:

∞⁰

set (2x+5)^1/x = y

take natural log of both sides and pull out exponent, yielding:

(1/x)ln(2x+5) = lny , another indeterminate form:

∞/∞

via Lôpetal’s Rule:

lny = lim_x->∞ (2/(2x+5)) / 1 = 0 so

lny = 0

e⁰ = y

1 = y

so lim_n->∞ a_n = 1

Question 11

what kind of expression is {a_n}?

Answer 11

a sequence

1, 2, 3, 4, 5, …

Question 12

a_n = (sin²n)/(√n)

Does {a_n} converge or diverge?

Answer 12

use squeeze therom

0 ≤ sin²n ≤ 1

0/(√n) ≤ (sin²n)/(√n) ≤ 1/(√n)

lim_n->∞ 1/(√n) = 0

lim_n->∞ = 0

Question 13

n! =

Answer 13

n! = n(n-1)(n-2)(n-3)•••3•2•1

Question 14

0! =

Answer 14

0! = 1

(by convention)

Question 15

7! =

Answer 15

7•6•5•4•3•2•1

Question 16

{n!} =

Answer 16

1, 2, 6, 24, 120, …

( 1!, 2!, 3!, 4!, 5!, … )

Question 17

how do you simplify a factorial like 6!/4! ?

Answer 17

6!/4! = (6•5•4•3•2•1)/(4•3•2•1)

which could be written:

(6•5•4!)/(4!)

factor out the factorial:

(6•5)(4!/4!)=

(6•5)(1)=

6•5

Question 18

(3n)! =

Answer 18

3n(3n-1)(3n-2)•••3•2•1

Question 19

{a_n} = {((-1)ⁿ)/n!)} =

define and explain how to find the limit

Answer 19

-1/1, 1/2, -1/3, 1/4, -1/5, …

cannot be rewritten as f(x)=(-1^x)/x! because x! is not the same as n! (x is not range restricted to natural numbers)

use squeeze therom instead of functional substitution and compare problem to 1/2ⁿ and -1/2ⁿ

f(x) = 1/2^x and g(x)=1/2^x

1/n! < 1/2ⁿ for n≥4 and -1/2ⁿ for n ≤ 4

-1/2ⁿ < (-1)ⁿ/n! < 1/2ⁿ for n ≥ 4

lim_n->∞ -1/2ⁿ = 0 and lim_n->∞ 1/2ⁿ = 0

so by squeeze therom lim_n->∞(-1)ⁿ/n! = 0

Question 20

list the heirarchy of growth of natural numbers from fastest to slowest, where n is a natural number and k is a constant

Answer 20

nⁿ

n!

kⁿ

n^k

√n

ln(n)

where k is a constant and n is a natural number

Question 21

does the series converge or diverge, and to where?

{√n/n³}

Answer 21

converges to 0

Question 22

does the series converge or diverge, and to where?

{2ⁿ/ln(n)}

Answer 22

diverges to ∞

Question 23

does the series converge or diverge, and to where?

{n!/nⁿ}

Answer 23

converges to 0

Question 24

does the series converge or diverge, and to where?

{(2ⁿ+1)/eⁿ}

Answer 24

converges to 0

Question 25

find an expression for the n^th term of a sequence:

3, 7, 11, 15, …

Answer 25

a_n = 4n-1

Question 26

find an expression for the n^th term of a sequence:

2, -1, 1/2, -1/4, 1/8, …

Answer 26

4(1/2)ⁿ(-1)ⁿ⁺¹

• or-
• (-1/2)^n-2
• or-
• 4(-1/2)ⁿ

(other answers possible)

Question 27

S_k =

(partial sums)

Answer 27

S_k = a₁+a₂+a₃+…a_k

Question 28

What is the definition of the sum of an infinite series?

Answer 28

The limit of its sequence of partial sums, provided the limit exists.

∑ _{<span>(</span>n=1}^∞) a_n = lim_k->∞ S_k

Question 29

∑ _(n=1^∞) 1/2ⁿ =

establish a pattern for generically determining the sequence of partial sums.

Answer 29

1/2+1/4+1/8+1/16+…

S₁ = 1/2

S₂ = 1/2+1/4 = 3/4

S₃ = 1/2+1/4+1/8 = 7/8

S₄ = 1/2+1/4/+1/8+1/16 = 15/16

S_k = (2^k-1)/(2^k)

so the sequence of partial sums:

{S^k} or {(2^k-1)/(2^k)}

so by the limit definition of the sequence of partial sums:

lim_k->∞(2^k-1)/2^k = 1

∑ ^(∞_n=1) 1/2ⁿ = 1

Question 30

∑ _(n=1^∞) 1/(n²-1) =

establish a pattern for generically determining the sequence of partial sums.

Answer 30

= 1/3+1/8+1/15+1/24+…

S₁ = 1/3

S₂ = 1/3+1/8 = 11/24

S₃ = 1/3+1/8+1/15 = 189/360

NO EASY PATTERN

Use partial fraction decomposition:

1(n²-1) = 1/((n-1)(n+1)) = A/(n-1)+B/(n+1)

1 = A(n+1)+B(n-1)

A = 1/2, B = -1/2

= ∑_(n=2^∞) ((1/2)/(n-1)+(-1/2)/(n+1)) (distribute:)

= ∑(n=2∞) (1/(2n-1)-1/(2n+1))

S₁ = 1/2-1/6

S₂ = (1/2-1/6)+(1/4-1/8)

S₃ = (1/2-1/6)+(1/4-1/8)+(1/6-1/10)

S₄ = (1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)

S₅ = (1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)+(1/10-1/14)

S_k = 1/2+1/4-1/(2(k+1))-1/(2(k+2))

lim_k->∞S_k = 1/2+1/4=3/4

so

∑ _(n=1^∞) 1/(n²-1) converges to 3/4

Question 31

AST

Answer 31

Alternating series test

If an alternating series satisfies:

1) lim_n->∞ a_n=0

2) a_n > a_n+1 > 0 (for all n > k, k is some positive integer)

then the series converges

if 2) fails, then AST does not apply

if 1) fails, tehn AST does not apply, and the limit of the sequence does not exist, so the n^th term test applies.

THE ALTERNATING SERIES TEST CANNOT PROVE DIVERGENCE

Question 33

Alternating Series Remainder

Answer 33

Suppose ∑(-1)ⁿ a_n is convergent by AST.

Then, if ∑(-1) a_n = S,

|S-S_N| < a_n+1 (where S_N is the N^th partial sum)