the ratio, root, and ratio comparison test

1. The Ratio Test and the Root test
We assume all series are positive series,
i.e. all terms in the series are positive unless
stated otherwise.

2. One fact of the convergent geometric series
The Ratio Test and the Root test
Σn=1
∞
rn is that
an+1
an
= rn+1
rn = r < 1.lim limn∞ n∞
We assume all series are positive series,
i.e. all terms in the series are positive unless
stated otherwise.

3. 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+1
an
= rn+1
rn = r < 1.lim limn∞ n∞
We assume all series are positive series,
i.e. all terms in the series are positive unless
stated otherwise.

4. 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+1
an
= rn+1
rn = r < 1.lim limn∞ n∞
We assume all series are positive series,
i.e. all terms in the series are positive unless
stated otherwise.
an+1
an
= r < 1, then converges,limn∞
Σn=1
anif
an+1
an
= r > 1, then diverges,limn∞
Σn=1
∞
anif
an+1
an
= 1, then the test is inconclusive.limn∞
if
Theorem ( Ratio Test) Given a series
∞
Σn=1
∞
an

5. The Ratio Test and the Root test
When we use the ratio test, it is easier to check
an+1* 1/an directly.

6. The Ratio Test and the Root test
Example A.
Let an =
3n
2n + n3
When we use the ratio test, it is easier to check
an+1* 1/an directly.

7. The Ratio Test and the Root test
Example A.
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

8. The Ratio Test and the Root test
Example A.
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

9. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ]

10. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ] (divide the top and bottom by 2n+1)

11. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ] (divide the top and bottom by 2n+1)
= 1
3
1 + (n+1)3/2n+1
1/2 + n3/2n+1[ ]

12. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ] (divide the top and bottom by 2n+1)
= 1
3
1 + (n+1)3/2n+1
1/2 + n3/2n+1[ ] as n∞,
0

13. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ] (divide the top and bottom by 2n+1)
= 1
3
1 + (n+1)3/2n+1
1/2 + n3/2n+1[ ] as n∞,
0
0

14. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ] (divide the top and bottom by 2n+1)
= 1
3
1 + (n+1)3/2n+1
1/2 + n3/2n+1[ ] as n∞, we get the limit
0
0
2
3 .

15. The Ratio Test and the Root test
Example A.
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
= 1
3
2n+1 + (n+1)3
2n + n3
[ ] (divide the top and bottom by 2n+1)
= 1
3
1 + (n+1)3/2n+1
1/2 + n3/2n+1[ ] as n∞, we get the limit
0
0
2
3 .
Hence the series converges.Σn=1
∞
3n
2n + n3

16. The Ratio Test and the Root test
Applying the ratio test to the harmonic series { },1
n
an+1* 1/an = n
n+1
1.we've:

17. The Ratio Test and the Root test
Applying the ratio test to the harmonic series { },1
n
an+1* 1/an = n
n+1
1.
Applying the ratio test to the 2-series { },
1
n2
an+1* 1/an = n2
(n+1)2 1.
we've:
we've:

18. The Ratio Test and the Root test
Applying the ratio test to the harmonic series { },1
n
an+1* 1/an = n
n+1
1.
Applying the ratio test to the 2-series { },
1
n2
an+1* 1/an = n2
(n+1)2 1.
The test results are the same, but the harmonic
series diverges and the 2-series converges.
we've:
we've:

19. The Ratio Test and the Root test
Applying the ratio test to the harmonic series { },1
n
an+1* 1/an = n
n+1
1.
Applying the ratio test to the 2-series { },
1
n2
an+1* 1/an = n2
(n+1)2 1.
The test results are the same, but the harmonic
series diverges and the 2-series converges.
So no conclusion may be drawn if the test limit is 1.
we've:
we've:

20. The Ratio Test and the Root test
Applying the ratio test to the harmonic series { },1
n
an+1* 1/an = n
n+1
1.
Applying the ratio test to the 2-series { },
1
n2
an+1* 1/an = n2
(n+1)2 1.
The test results are the same, but the harmonic
series diverges and the 2-series converges.
So no conclusion may be drawn if the test limit is 1.
Another fact of the convergent geometric series is that
an
= = r < 1.lim limn∞ n∞
n
rn
n
we've:
we've:

21. The Ratio Test and the Root test
Applying the ratio test to the harmonic series { },1
n
an+1* 1/an = n
n+1
1.
Applying the ratio test to the 2-series { },
1
n2
an+1* 1/an = n2
(n+1)2 1.
The test results are the same, but the harmonic
series diverges and the 2-series converges.
So no conclusion may be drawn if the test limit is 1.
Another fact of the convergent geometric series is that
an
= = r < 1.lim limn∞ n∞
n
rn
n
we've:
we've:
This observation leads to the root test which
checks the limit of the n´th root ofan.

22. The Ratio Test and the Root test
Theorem (Root Test)
, let an
= r,limn∞
n
ΣanGiven the series
∞

23. The Ratio Test and the Root test
Theorem (Root Test)
, let an
= r,limn∞
n
if r < 1, then converges,Σan
Σan
if r = 1, the test failed.
Given the series
∞
if r > 1, then diverges,Σan

24. The Ratio Test and the Root test
A useful fact, when applying the root test, is that
lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule.
n∞
Theorem (Root Test)
, let an
= r,limn∞
n
if r < 1, then converges,Σan
Σan
if r = 1, the test failed.
Given the series
∞
if r > 1, then diverges,Σan

25. The Ratio Test and the Root test
A useful fact, when applying the root test, is that
lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule.
n∞
Example B.
3n
n3
Theorem (Root Test)
, let an
= r,limn∞
n
if r < 1, then converges,Σan
Σan
if r = 1, the test failed.
Given the series
∞
if r > 1, then diverges,Σan
Let an =

26. The Ratio Test and the Root test
A useful fact, when applying the root test, is that
lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule.
n∞
Example B.
3n
n3
, the n'th root of an = n3/3n
n
= n3/n/3.
Theorem (Root Test)
, let an
= r,limn∞
n
if r < 1, then converges,Σan
Σan
if r = 1, the test failed.
Given the series
∞
if r > 1, then diverges,Σan
Let an =

27. The Ratio Test and the Root test
A useful fact, when applying the root test, is that
lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule.
n∞
Example B.
3n
n3
, the n'th root of an = n3/3n
n
= n3/n/3.
Since lim n3/n = 1, so the lim (an)1/n = 1/3 < 1.n∞
Theorem (Root Test)
, let an
= r,limn∞
n
if r < 1, then converges,Σan
Σan
if r = 1, the test failed.
Given the series
∞
if r > 1, then diverges,Σan
Let an =

28. The Ratio Test and the Root test
A useful fact, when applying the root test, is that
lim (nk)1/n = 1 which may be verified by the
L'Hopital's Rule.
n∞
Example B.
3n
n3
, the n'th root of an = 
Hence the series converges.Σn=1
∞
3n
n3
n3/3n
n
= n3/n/3.
Since lim n3/n = 1, so the lim (an)1/n = 1/3 < 1.n∞
Theorem (Root Test)
, let an
= r,limn∞
n
if r < 1, then converges,Σan
Σan
if r = 1, the test failed.
Given the series
∞
if r > 1, then diverges,Σan
Let an =

29. The Ratio ComparisonTest

30. The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
The Ratio ComparisonTest
Σn=1
∞
an

31. The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

32. The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
If a series is the constant multiple of another series,
then the two series behave the same.
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

33. The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
If converges then the series converges.Σn=1
∞
an Σn=1
∞
can
If a series is the constant multiple of another series,
then the two series behave the same. Specifically,
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

34. The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
If converges then the series converges.Σn=1
∞
an Σn=1
∞
can
If diverges then the series diverges, c = 0.Σn=1
∞
an Σn=1
∞
can
If a series is the constant multiple of another series,
then the two series behave the same. Specifically,
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

35. We can draw the same conclusion if two series are
close enough to be the multiple of each other.
The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
If converges then the series converges.Σn=1
∞
an Σn=1
∞
can
If diverges then the series diverges, c = 0.Σn=1
∞
an Σn=1
∞
can
If a series is the constant multiple of another series,
then the two series behave the same. Specifically,
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

36. We can draw the same conclusion if two series are
close enough to be the multiple of each other.
The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
If converges then the series converges.Σn=1
∞
an Σn=1
∞
can
If diverges then the series diverges, c = 0.Σn=1
∞
an Σn=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 = 0n∞
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

37. We can draw the same conclusion if two series are
close enough to be the multiple of each other.
The ratio and root tests for the convergence of
attempt to make conclusion directly from {an}.
But if needed, we may apply the test to a simpler
surrogate series bn to test for convergence of .
The Ratio ComparisonTest
If converges then the series converges.Σn=1
∞
an Σn=1
∞
can
If diverges then the series diverges, c = 0.Σn=1
∞
an Σn=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 and two series aren∞
almost-multiple if {an}, {bn} are almost-multiple.
Σn=1
∞
an
Σn=1
∞
Σn=1
∞
an

38. The Ratio ComparisonTest

39. The Ratio ComparisonTest
Theorem (Limit or Ratio Comparison)
If two series and are almost-multiple of each
other, then they behave the same.
Σn=1
∞
an Σn=1
∞
bn

40. The Ratio ComparisonTest
Theorem (Limit or Ratio Comparison)
If two series and are almost-multiple of each
other, then they behave the same.
Hence converges if and only if converges.Σ
∞
an Σ
∞
bn
Remark: If lim and converges,
bn
an
= 0n∞
Σn=1
∞
an
Σn=1
∞
bn converges also.
Σn=1
∞
an Σn=1
∞
bn
then
n=1 n=1

41. The Ratio ComparisonTest
Remark: If lim and converges,
bn
an
= 0n∞
Σn=1
∞
an
Σn=1
∞
bn converges also.
Most of the series are given as fractional terms and
their behavior is determined by the dominant-terms
of the numerator and the denominator.
then
Theorem (Limit or Ratio Comparison)
If two series and are almost-multiple of each
other, then they behave the same.
Hence converges if and only if converges.Σ
∞
an Σ
∞
bn
Σn=1
∞
an Σn=1
∞
bn
n=1 n=1

42. The Ratio ComparisonTest
Remark: If lim and converges,
bn
an
= 0n∞
Σn=1
∞
an
Σn=1
∞
bn converges also.
Most of the series are given as fractional terms and
their behavior is determined by the dominant-terms
of the numerator and the denominator.
With the limit comparison theorem, we may drop
the irrelevant terms in an, then do the root or ratio tests
to simplify the algebra.
then
Theorem (Limit or Ratio Comparison)
If two series and are almost-multiple of each
other, then they behave the same.
Hence converges if and only if converges.Σ
∞
an Σ
∞
bn
Σn=1
∞
an Σn=1
∞
bn
n=1 n=1

43. 4n + 2
n2 + n – 1
The Ratio ComparisonTest
Example C.
Let an =

44. 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Example C.
Let an =

45. 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Let bn = n
n2 , and use the limit comparison theorem.
Example C.
Let an =

46. 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Let bn = n
n2 , and use the limit comparison theorem.
lim
bn
ann∞
= lim n
n2 4n + 2
n2 + n – 1
n∞
Example C.
Let an =

47. 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Let bn = n
n2 , and use the limit comparison theorem.
lim
bn
ann∞
= lim n
n2 4n + 2
n2 + n – 1
n∞
= lim n2
n
4n + 2
n2 + n – 1n∞
= 1/4
Example C.
Let an =

48. 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Let bn = n
n2 , and use the limit comparison theorem.
lim
bn
ann∞
= lim n
n2 4n + 2
n2 + n – 1
n∞
= lim n2
n
4n + 2
n2 + n – 1n∞
Hence {an} and {bn} are almost-multiple of each other.
Example C.
Let an =
= 1/4

49. 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Let bn = n
n2 , and use the limit comparison theorem.
lim
bn
ann∞
= lim n
n2 4n + 2
n2 + n – 1
n∞
= lim n2
n
4n + 2
n2 + n – 1n∞
Hence {an} and {bn} are almost-multiple of each other.
Σn=1
∞
bn diverges,= Σn=1
∞
1
n
Example C.
Let an =
= 1/4

50. Example C.
Let an = 4n + 2
n2 + n – 1
, the dominant term in the
The Ratio ComparisonTest
numerator is 4n, and for the denominator is n2.
Let bn = n
n2 , and use the limit comparison theorem.
lim
bn
ann∞
= lim n
n2 4n + 2
n2 + n – 1
n∞
= lim n2
n
4n + 2
n2 + n – 1n∞
Hence {an} and {bn} are almost-multiple of each other.
Σn=1
∞
bn diverges, hence= Σn=1
∞
1
n Σn=1
∞
an diverges also.
= 1/4

51. n7 + 2
3n2 + n – 1
The Ratio ComparisonTest
Example D.
Let an =

52. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Example D.
Let an =

53. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
Example D.
Let an =

54. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
Example D.
Let an =

55. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
Example D.
Let an =

56. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
= lim n2
1 + 2/n7
3n2 + n – 1n∞
*
Example D.
Let an =

57. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
= lim n2
1 + 2/n7
3n2 + n – 1n∞
*
0
Example D.
Let an =
, for the denominator is

58. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
= lim n2
1 + 2/n7
3n2 + n – 1n∞
* = 1
3
0
Example D.
Let an =

59. n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
= lim n2
1 + 2/n7
3n2 + n – 1n∞
* = 1
3
0
Hence {an} and {bn} are almost-multiple of each other.
Example D.
Let an =

60. Example D.
Let an =
n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
= lim n2
1 + 2/n7
3n2 + n – 1n∞
* = 1
3
0
Hence {an} and {bn} are almost-multiple of each other.
Σn=1
∞
bn converges,= Σn=1
∞
1
n3/2

61. Example D.
Let an =
n7 + 2
3n2 + n – 1 , the dominant term in the
The Ratio ComparisonTest
numerator is 3n2, and for the denominator is n7/2.
Let bn = n2
n7/2, and use the limit comparison theorem.
lim
bn
ann∞
= lim n2
n7/2
n7 + 2
3n2 + n – 1n∞
= lim n2
n7
n7 + 2
3n2 + n – 1n∞
= lim n2
1 + 2/n7
3n2 + n – 1n∞
* = 1
3
0
Hence {an} and {bn} are almost-multiple of each other.
Σn=1
∞
bn converges, hence= Σn=1
∞
1
n3/2 Σn=1
∞
an converges also.