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25 the ratio, root, and ratio comparison test x

The document discusses two tests for determining if an infinite series converges or diverges: the ratio test and the root test. The ratio test checks if the limit of the ratio of successive terms is less than 1, equal to 1, or greater than 1, indicating convergence, inconclusiveness, or divergence respectively. The root test checks if the limit of the nth root of terms is less than, equal to, or greater than 1 with similar implications. Examples are provided to demonstrate how to apply each test. It is also noted that if a series cannot be directly tested, an alternative surrogate series may be used instead if their behaviors are the same.

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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.
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.
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.
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
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 A. 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 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
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
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 [ ]
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)
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[ ]
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
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
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 .
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
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:
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:
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:
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:
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:
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.
The Ratio Test and the Root test Theorem (Root Test) , let an = r,limn∞ n ΣanGiven the series ∞
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
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
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 =
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 =
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 =
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 =
The Ratio ComparisonTest
The ratio and root tests for the convergence of attempt to make conclusion directly from {an}. The Ratio ComparisonTest Σn=1 ∞ an
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
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
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
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
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
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
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
The Ratio ComparisonTest
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
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
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
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
4n + 2 n2 + n – 1 The Ratio ComparisonTest 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. 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. 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∞ 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∞ = 1/4 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. Example C. Let an = = 1/4
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
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
n7 + 2 3n2 + n – 1 The Ratio ComparisonTest 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. 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. 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∞ 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∞ 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∞ * 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∞ * 0 Example D. Let an = , for the denominator is
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 =
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 =
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
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.

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25 the ratio, root, and ratio comparison test x

  • 1.
    The Ratio Testand 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 ofthe 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 Testand the Root test When we use the ratio test, it is easier to check an+1* 1/an directly.
  • 6.
    The Ratio Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand the Root test Theorem (Root Test) , let an = r,limn∞ n ΣanGiven the series ∞
  • 23.
    The Ratio Testand 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 Testand 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 Testand 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 Testand 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 Testand 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 Testand 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.
  • 30.
    The ratio androot tests for the convergence of attempt to make conclusion directly from {an}. The Ratio ComparisonTest Σn=1 ∞ an
  • 31.
    The ratio androot 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 androot 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 androot 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 androot 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 drawthe 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 drawthe 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 drawthe 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.
  • 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.