The Ratio Test and the Root test
26 the ratio and root test
Jun 27, 2015
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The Ratio Test and the Root test
The Ratio Test and the Root testWe shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞rn is that
an+1an
= rn+1
rn = r < 1.lim limn∞ n∞
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
In other words, the limit of the ratio of the terms exists and is less than 1.
One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞rn is that
an+1an
= rn+1
rn = r < 1.lim limn∞ n∞
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
In other words, the limit of the ratio of the terms exists and is less than 1.
One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞rn is that
an+1an
= rn+1
rn = r < 1.lim limn∞ n∞
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
Theorem (Ratio Test):
In other words, the limit of the ratio of the terms exists and is less than 1.
One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞rn is that
an+1an
= rn+1
rn = r < 1.lim limn∞ n∞
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
Theorem (Ratio Test): Given a series an+1an
= r < 1, then converges. limn∞
Σn=1
∞anIf
Σn=1
∞an
In other words, the limit of the ratio of the terms exists and is less than 1.
One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞rn is that
an+1an
= rn+1
rn = r < 1.lim limn∞ n∞
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
Theorem (Ratio Test): Given a series an+1an
= r < 1, then converges. limn∞
Σn=1
∞anIf
an+1an
= r > 1, then diverges. limn∞
Σn=1
∞anIf
Σn=1
∞an
In other words, the limit of the ratio of the terms exists and is less than 1.
One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞rn is that
an+1an
= rn+1
rn = r < 1.lim limn∞ n∞
We shall assume all series are positive series, i.e. all terms in the series are positive unless stated otherwise.
Theorem (Ratio Test): Given a series an+1an
= r < 1, then converges. limn∞
Σn=1
∞anIf
an+1an
= r > 1, then diverges. limn∞
Σn=1
∞anIf
an+1an
= 1, then the test is inconclusive.limn∞
If
Σn=1
∞an
The Ratio Test and the Root test
When we use the ratio test, it is easier to check an+1* 1/an directly.
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ]
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ] (divide the top and bottom by 2n+1)
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ] (divide the top and bottom by 2n+1)
= 13
1 + (n+1)3/2n+1
1/2 + n/2n+1[ ]
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ] (divide the top and bottom by 2n+1)
= 13
1 + (n+1)3/2n+1
1/2 + n/2n+1[ ] as n0, 0
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ] (divide the top and bottom by 2n+1)
= 13
1 + (n+1)3/2n+1
1/2 + n/2n+1[ ] as n0, 0
0
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ] (divide the top and bottom by 2n+1)
= 13
1 + (n+1)3/2n+1
1/2 + n/2n+1[ ] as n0, we get the limit 0
0
23 .
The Ratio Test and the Root test
Example:
Let an = 3n
2n + n3
When we use the ratio test, it is easier to check an+1* 1/an directly.
, then an+1 = 3n+1
2n+1 + (n+1)3
So an+1 * 1/an = 3n+1
2n+1 + (n+1)3 3n
2n + n3
= 13
2n+1 + (n+1)3
2n + n3[ ] (divide the top and bottom by 2n+1)
= 13
1 + (n+1)3/2n+1
1/2 + n/2n+1[ ] as n0, we get the limit 0
0
23 .
Hence the series converges.Σn=1
∞
3n
2n + n3
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, 1n
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, 1n2
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The harmonic series diverges,
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The harmonic series diverges, but the 2-seriesconverges.
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The harmonic series diverges, but the 2-seriesconverges. Hence no conclusion may be drawn if the limit is 1.
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The harmonic series diverges, but the 2-seriesconverges. Hence no conclusion may be drawn if the limit is 1.
Another fact of the convergent geometric series is that
an= = r < 1.lim lim
n∞ n∞
nrnn
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The harmonic series diverges, but the 2-seriesconverges. Hence no conclusion may be drawn if the limit is 1.
Another fact of the convergent geometric series is that
an= = r < 1.lim lim
n∞ n∞
nrnn
Theorem (Root Test):
The Ratio Test and the Root test
Applying ratio test to the harmonic series { }, we've 1n
an+1*1/an = nn+1 1.
If we apply ratio test to the 2-series { }, we've 1n2
an+1*1/an = n2
(n+1)2 1.
The harmonic series diverges, but the 2-seriesconverges. Hence no conclusion may be drawn if the limit is 1.
Another fact of the convergent geometric series is that
an= = r < 1.lim lim
n∞ n∞
nrnn
Theorem
If an= r < 1,lim
n∞
nthen converges. Σ
n=1
∞an
(Root Test): Given a series Σn=1
∞an
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
Example:
Let an = 3nn3
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
Example:
Let an = 3nn3
, the n'th root of an = n3/3nn
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
Example:
Let an = 3nn3
, the n'th root of an = n3/3nn
= n3/n/3
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
Example:
Let an = 3nn3
, the n'th root of an = n3/3nn
= n3/n/3
Since lim n3/n = 1, n∞
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
Example:
Let an = 3nn3
, the n'th root of an = n3/3nn
= n3/n/3
Since lim n3/n = 1, so the lim (an)1/n = 1/3 < 1.n∞
The Ratio Test and the Root test
If an= r > 1,lim
n∞
nthen diverges. Σ
n=1
∞an
If an= 1,lim
n∞
nthen the test failed.
A useful fact to recall concerning the root test is that lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule. n∞
Example:
Let an = 3nn3
, the n'th root of an =
Hence the series converges.Σn=1
∞
3nn3
n3/3nn
= n3/n/3
Since lim n3/n = 1, so the lim (an)1/n = 1/3 < 1.n∞
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