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Page 1: 22 sequences

Sequences

Page 2: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Sequences

Page 3: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.

Sequences

Page 4: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.1, 4, 9, 16, 25, …is the sequence of square numbers.

Sequences

Page 5: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.1, 4, 9, 16, 25, …is the sequence of square numbers.5, -2, , e2, -110, …is a sequence without an obvious pattern.

Sequences

Page 6: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.1, 4, 9, 16, 25, …is the sequence of square numbers.5, -2, , e2, -110, …is a sequence without an obvious pattern.

as {f(n)}n=1.

Sequences

One way to describe a sequence is to give the formula for the general term of the sequence, that is

Page 7: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.1, 4, 9, 16, 25, …is the sequence of square numbers.5, -2, , e2, -110, …is a sequence without an obvious pattern.

as {f(n)}n=1. We also write f(n) as an.

Sequences

One way to describe a sequence is to give the formula for the general term of the sequence, that is

Page 8: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.1, 4, 9, 16, 25, …is the sequence of square numbers.5, -2, , e2, -110, …is a sequence without an obvious pattern.

as {f(n)}n=1. We also write f(n) as an.

Sequences

One way to describe a sequence is to give the formula for the general term of the sequence, that is

Example:A. The sequence {3n + 1}n=1 =

Page 9: 22 sequences

A sequence is an ordered list of infinitely many numbers that may or may not have a pattern.

Example:1, 3, 5, 7, 9,… is the sequence of odd numbers.1, 4, 9, 16, 25, …is the sequence of square numbers.5, -2, , e2, -110, …is a sequence without an obvious pattern.

as {f(n)}n=1. We also write f(n) as an.

Sequences

One way to describe a sequence is to give the formula for the general term of the sequence, that is

Example:A. The sequence {3n + 1}n=1 = {4, 7, 10, …} ∞

Page 10: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

Sequences

Page 11: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

A sequence with alternating positive and negative terms is called an alternating sequence.

Sequences

Page 12: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

A sequence with alternating positive and negative terms is called an alternating sequence.The formulas for alternating sequences contain the multiplier of (–1)n. which switches the sign.

Sequences

Page 13: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

A sequence with alternating positive and negative terms is called an alternating sequence.The formulas for alternating sequences contain the multiplier of (–1)n. which switches the sign.

Example C. The sequence {(–1)n(2n – 1)}n=1 is the alternating sequence –1, 3, –5, 7, –9, …

Sequences

Page 14: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

A sequence with alternating positive and negative terms is called an alternating sequence.The formulas for alternating sequences contain the multiplier of (–1)n. which switches the sign.

Example C. The sequence {(–1)n(2n – 1)}n=1 is the alternating sequence –1, 3, –5, 7, –9, …

D. Give the formula for the sequence , , , , ...-49

23

627

-881

Sequences

Page 15: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

A sequence with alternating positive and negative terms is called an alternating sequence.The formulas for alternating sequences contain the multiplier of (–1)n. which switches the sign.

Example C. The sequence {(–1)n(2n – 1)}n=1 is the alternating sequence –1, 3, –5, 7, –9, …

D. Give the formula for the sequence , , , , ...-49

23

627

-881

The numerators are 2n, the denominators are 3n andthe signs alternates, so its {(-1)n-12n/3n}n=1

. ∞

Sequences

Page 16: 22 sequences

B. The sequence of odd numbers 1, 3, , …may be given as {2n – 1}, n =1, 2, ...It also may be given as {2n +1}, n = 0, 1, 2, …

A sequence with alternating positive and negative terms is called an alternating sequence.The formulas for alternating sequences contain the multiplier of (–1)n. which switches the sign.

Example C. The sequence {(–1)n(2n – 1)}n=1 is the alternating sequence –1, 3, –5, 7, –9, …

D. Give the formula for the sequence , , , , ...-49

23

627

-881

The numerators are 2n, the denominators are 3n andthe signs alternates, so its {(-1)n-12n/3n}n=1

. ∞

Sequences

Page 17: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists.

x∞x∞

Sequences

Page 18: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 19: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 20: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x

x∞n∞

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 21: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x )1/x

x∞n∞ x∞

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 22: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x ) = lim eLn(x)/x1/x

x∞n∞ x∞ x∞

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 23: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x ) = lim eLn(x)/x1/x

x∞n∞ x∞ x∞

0

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 24: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x ) = lim eLn(x)/x = e0 = 11/x

x∞n∞ x∞ x∞

0

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 25: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x ) = lim eLn(x)/x = e0 = 11/x

x∞n∞ x∞ x∞

0

Hence an = { n1/n } is a convergent (CG) sequence.

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 26: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x ) = lim eLn(x)/x = e0 = 11/x

x∞n∞ x∞ x∞

0

n∞ n∞

The Sandwich Theorem:If 0 ≤ an ≤ bn for all sufficiently large* n and that lim bn = 0, then lim an = 0.

Hence an = { n1/n } is a convergent (CG) sequence.

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 27: 22 sequences

Limit of a sequence lim f(n) is just lim f(x), if it exists. x∞x∞

Sequences

Example: Let an = { n1/n }, find lim an if it exists.x∞

lim n1/n = lim x1/x = lim eLn(x ) = lim eLn(x)/x = e0 = 11/x

x∞n∞ x∞ x∞

0

n∞ n∞

* “Sufficiently large n" means that "for all the numbers n that are greater than certain fixed number".

The Sandwich Theorem:If 0 ≤ an ≤ bn for all sufficiently large* n and that lim bn = 0, then lim an = 0.

Hence an = { n1/n } is a convergent (CG) sequence.

We say the sequence converges (CG) if the limit is finite, we say it diverges (DG) if the limit is ±∞ or if it doesn't exist.

Page 28: 22 sequences

SequencesExample: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 29: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn,

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 30: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2n

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 31: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 32: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 33: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 34: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 35: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120.

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 36: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120. (We define 0! = 1)

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 37: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120. (We define 0! = 1)

Example: Justify that lim 1/n! = 0 as n∞

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Page 38: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120. (We define 0! = 1)

Example: Justify that lim 1/n! = 0 as n∞

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Compare {1/n!} with {1/n},

Page 39: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120. (We define 0! = 1)

Example: Justify that lim 1/n! = 0 as n∞1/n > 1/n! > 0 for n > 3.

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Compare {1/n!} with {1/n},

Page 40: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120. (We define 0! = 1)

Example: Justify that lim 1/n! = 0 as n∞1/n > 1/n! > 0 for n > 3.

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Compare {1/n!} with {1/n},Since lim 1/n = 0

Page 41: 22 sequences

Sequences

Need to find a number c so that if n > c we’ve an > bn, i.e. n2 – 10 > 2nn2 – 2n – 10 > 0(n – 5)(n + 3) > 0. From this inequality, we see that if we set c = 5 then for n > c we’ll have that an > bn.

Recall for n = 0, 1, 2,.., the numbern! = n(n – 1)(n – 2)…3*2*1 is called the n factorial.For example 5! = 5*4*3*2*1 = 120. (We define 0! = 1)

Example: Justify that lim 1/n! = 0 as n∞1/n > 1/n! > 0 for n > 3.

Example: Given that {an} = {n2 – 15} and {bn} = {2n} for n = 1, 2, .. show that an > bn for sufficiently large n.

Compare {1/n!} with {1/n},Since lim 1/n = 0 so lim 1/n! = 0 as n ∞.

Page 42: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

Page 43: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

Compare bn = {1011/n} to an = {10n/n!}.

Page 44: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

Compare bn = {1011/n} to an = {10n/n!}.

Page 45: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Page 46: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

Page 47: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

Page 48: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!= 10 * 10….. 10*10*10*10…10

n *(n-1)….11*10* 9 * 8…..1

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

Page 49: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!= 10 * 10….. 10*10*10*10…10

n *(n-1)….11*10* 9 * 8…..1

1

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

Page 50: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!= 10 * 10….. 10*10*10*10…10

n *(n-1)….11*10* 9 * 8…..1

10101

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

Page 51: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!= 10 * 10….. 10*10*10*10…10

n *(n-1)….11*10* 9 * 8…..1

10101

< 1011

n

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

Page 52: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!= 10 * 10….. 10*10*10*10…10

n *(n-1)….11*10* 9 * 8…..1

10101

< 1011

n

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

0n∞

Page 53: 22 sequences

Sequencesn∞

Example: Justify that lim 10n/n! = 0

We will show that bn > an (> 0) for sufficiently large n,

For if n > 10 we note that

10n

n!= 10 * 10….. 10*10*10*10…10

n *(n-1)….11*10* 9 * 8…..1

10101

< 1011

n

Hence for large n, 1011/n > 10n/n!, and the fact 1011/n 0 implies lim10n/n! = 0

n∞

Compare bn = {1011/n} to an = {10n/n!}.

and since that bn0 we must have that an0.

Specifically we claim that bn > an for n > 10

0

n∞