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. ∑ _(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)