27 the ratio comparison test

The Ratio ComparisonTest

We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.

The Ratio ComparisonTest

We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.

The Ratio ComparisonTest

If a series is the constant multiple of another series, then the two series behave the same.

We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.

The Ratio ComparisonTest

If converges then the series converges.

Σn=1

∞an Σ

n=1

∞can

If a series is the constant multiple of another series, then the two series behave the same. Specifically,

We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.

The Ratio ComparisonTest

If converges then the series converges.

Σn=1

∞an Σ

n=1

∞can

If diverges then the series diverges, c = 0.

Σn=1

∞an Σ

n=1

∞can

If a series is the constant multiple of another series, then the two series behave the same. Specifically,

These facts may be generalized.

We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.

The Ratio ComparisonTest

If converges then the series converges.

Σn=1

∞an Σ

n=1

∞can

If diverges then the series diverges, c = 0.

Σn=1

∞an Σ

n=1

∞can

If a series is the constant multiple of another series, then the two series behave the same. Specifically,

These facts may be generalized.

We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.

The Ratio ComparisonTest

If converges then the series converges.

Σn=1

∞an Σ

n=1

∞can

If diverges then the series diverges, c = 0.

Σn=1

∞an Σ

n=1

∞can

We say two sequences {an}, {bn} are almost-multiple

of each other if lim

If a series is the constant multiple of another series, then the two series behave the same. Specifically,

bn

an = c = 0.

n∞

The Ratio ComparisonTest

Theorem (Limit Comparison): Given two series that are almost-multiple of each other, then they behave the same.

The Ratio ComparisonTest

Theorem (Limit Comparison): Given two sequences that are almost-multiple of each other, then they behave the same.

That is, converges if and if only if converges. Σn=1

∞an Σ

n=1

∞bn

The Ratio ComparisonTest

Theorem (Limit Comparison): Given two series that are almost-multiple of each other, then they behave the same.

That is, converges if and only if converges. Σn=1

∞an Σ

n=1

∞bn

and diverges if and only if diverges. Σn=1

∞an Σ

n=1

∞bn

The Ratio ComparisonTest

Theorem (Limit Comparison): Given two series that are almost-multiple of each other, then they behave the same.

That is, converges if and only if converges. Σn=1

∞an Σ

n=1

∞bn

and diverges if and only if diverges. Σn=1

∞an Σ

n=1

∞bn

Remark: If lim and converges, thenbn

an = 0

n∞Σn=1

∞an

Σn=1

∞bn converges also.

The Ratio ComparisonTest

Theorem (Limit Comparison): Given two series that are almost-multiple of each other, then they behave the same.

That is, converges if and only if converges. Σn=1

∞an Σ

n=1

∞bn

and diverges if and only if diverges. Σn=1

∞an Σ

n=1

∞bn

Remark: If lim and converges, thenbn

an = 0

n∞Σn=1

∞an

Σn=1

∞bn converges also.

Most of the series are given as fractional terms. Behavior of fractions are determined by the dominate-terms of the numerator and the denominator.

The Ratio ComparisonTest

Theorem (Limit Comparison): Given two series that are almost-multiple of each other, then they behave the same.

That is, converges if and only if converges. Σn=1

∞an Σ

n=1

∞bn

and diverges if and only if diverges. Σn=1

∞an Σ

n=1

∞bn

Remark: If lim and converges, thenbn

an = 0

n∞Σn=1

∞an

Σn=1

∞bn converges also.

Most of the series are given as fractional terms. Behavior of fractions are determined by the dominate-terms of the numerator and the denominator. We investigate the series by dropping the irrelavent terms in the fraction and use limit comparison test.

Example:

Let an = 4n + 2n2 + n – 1

The Ratio ComparisonTest

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

limbn

an n∞= lim n

n2 4n + 2n2 + n – 1

n∞

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

limbn

an n∞= lim n

n2 4n + 2n2 + n – 1

n∞

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

limbn

an n∞= lim n

n2 4n + 2n2 + n – 1

n∞

= lim n2

n4n + 2

n2 + n – 1 n∞= 4

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

limbn

an n∞= lim n

n2 4n + 2n2 + n – 1

n∞

= lim n2

n4n + 2

n2 + n – 1 n∞= 4

Hence {an} and {bn} are almost-multiple of each other.

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

limbn

an n∞= lim n

n2 4n + 2n2 + n – 1

n∞

= lim n2

n4n + 2

n2 + n – 1 n∞= 4

Hence {an} and {bn} are almost-multiple of each other.

Σn=1

∞bn diverges,= Σ

n=1

∞ 1n

Example:

Let an = 4n + 2n2 + n – 1

, the dominate terms in the

The Ratio ComparisonTest

numerator is 4n, for the denominator is n2.

Let bn = nn2 , and use the limit comparison theorem.

limbn

an n∞= lim n

n2 4n + 2n2 + n – 1

n∞

= lim n2

n4n + 2

n2 + n – 1 n∞= 4

Hence {an} and {bn} are almost-multiple of each other.

Σn=1

∞bn diverges, hence = Σ

n=1

∞ 1n Σ

n=1

∞an diverges also.

Example:

Let an = n7 + 2

3n2 + n – 1

The Ratio ComparisonTest

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

= lim n2

1 + 2/n73n2 + n – 1 n∞ *

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

= lim n2

1 + 2/n73n2 + n – 1 n∞ *

0

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

= lim n2

1 + 2/n73n2 + n – 1 n∞ * = 1

3

0

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

= lim n2

1 + 2/n73n2 + n – 1 n∞ * = 1

3

0

Hence {an} and {bn} are almost-multiple of each other.

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

= lim n2

1 + 2/n73n2 + n – 1 n∞ * = 1

3

0

Hence {an} and {bn} are almost-multiple of each other.

Σn=1

∞bn converges, = Σ

n=1

∞ 1n3/2

Example:

Let an = n7 + 2

3n2 + n – 1 , the dominate terms in the

The Ratio ComparisonTest

numerator is 3n2, for the denominator is n7/2.

Let bn = n2

n7/2, and use the limit comparison theorem.

limbn

an n∞= lim n2

n7/2n7 + 23n2 + n – 1 n∞

= lim n2

n7

n7 + 23n2 + n – 1 n∞

= lim n2

1 + 2/n73n2 + n – 1 n∞ * = 1

3

0

Hence {an} and {bn} are almost-multiple of each other.

Σn=1

∞bn converges, hence = Σ

n=1

∞ 1n3/2 Σ

n=1

∞an converges also.