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The Ratio Comparison Test
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
The Ratio Comparison Test
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
The Ratio Comparison Test
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 Comparison Test
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 Comparison Test
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 Comparison Test
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 Comparison Test
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 Comparison Test Theorem (Limit Comparison): Given two series that are almost-multiple of each other, then they behave the same.
The Ratio Comparison Test 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 Comparison Test 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 Comparison Test 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 Comparison Test 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 + 2
n2 + n – 1
The Ratio Comparison Test
Example:Let an = 4n + 2
n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 4n, for the denominator is n2.
Example:Let an = 4n + 2
n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 4n, for the denominator is n2.
Let bn = nn2 , and use the limit comparison theorem.
Example:Let an = 4n + 2
n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 4n, for the denominator is n2.
Let bn = nn2 , and use the limit comparison theorem.
limbnan n∞ = lim n
n2 4n + 2n2 + n – 1
n∞
Example:Let an = 4n + 2
n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 4n, for the denominator is n2.
Let bn = nn2 , and use the limit comparison theorem.
limbnan n∞ = lim n
n2 4n + 2n2 + n – 1
n∞
= 1/4
Hence {an} and {bn} are almost-multiple of each other.
Example:Let an = 4n + 2
n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 4n, for the denominator is n2.
Let bn = nn2 , and use the limit comparison theorem.
limbnan n∞ = lim n
n2 4n + 2n2 + n – 1
n∞
= 1/4
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 + 23n2 + n – 1
The Ratio Comparison Test
Example:Let an =
n7 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2.
Example:Let an =
n7 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
Example:Let an =
n7 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan n∞ = lim n2
n7/2n7 + 23n2 + n – 1 n∞
Example:Let an =
n7 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan n∞ = lim n2
n7/2n7 + 23n2 + n – 1 n∞
= lim n2
n7
n7 + 23n2 + n – 1 n∞
Example:Let an =
n7 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan 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 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan 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 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan 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 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan 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 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan 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 + 23n2 + n – 1 , the dominate terms in the
The Ratio Comparison Test
numerator is 3n2, for the denominator is n7/2. Let bn = n2
n7/2, and use the limit comparison theorem.
limbnan 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.
