Lecture 26 : Comparison Test P P I Comparison Test Suppose That an and Bn Are Series with Positive Terms

I We will of course make use of our knowledge of p-series and geometric series.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

∞ X n−1 ar converges if |r| < 1, diverges if |r| ≥ 1. n=1 P P I Comparison Test Suppose that an and bn are series with positive terms. P P (i) If bn is convergent and an ≤ bn for all n, than an is also convergent. P P (ii) If bn is divergent and an ≥ bn for all n, then an is divergent.

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In this section, as we did with improper integrals, we see how to compare a series (with Positive terms) to a well known series to determine if it converges or diverges.

Annette Pilkington Lecture 26 : Comparison Test P P I Comparison Test Suppose that an and bn are series with positive terms. P P (i) If bn is convergent and an ≤ bn for all n, than an is also convergent. P P (ii) If bn is divergent and an ≥ bn for all n, then an is divergent.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Comparison Test

In this section, as we did with improper integrals, we see how to compare a series (with Positive terms) to a well known series to determine if it converges or diverges.

I We will of course make use of our knowledge of p-series and geometric series.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

∞ X n−1 ar converges if |r| < 1, diverges if |r| ≥ 1. n=1

In this section, as we did with improper integrals, we see how to compare a series (with Positive terms) to a well known series to determine if it converges or diverges.

I We will of course make use of our knowledge of p-series and geometric series.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

Annette Pilkington Lecture 26 : Comparison Test 2−1/n I First we check that an > 0 –> true since n3 > 0 for n ≥ 1. √ We have 21/n = n 2 > 1 for n ≥ 1. Therefore 2−1/n = √1 < 1 for n ≥ 1. I n 2 2−1/n 1 I Therefore n3 < n3 for n > 1. P∞ 1 I Since n=1 n3 is a p-series with p > 1, it converges. P∞ 1 I Comparing the above series with n=1 n3 , we can conclude that P∞ 2−1/n P∞ 2−1/n P∞ 1 n=1 n3 also converges and n=1 n3 ≤ n=1 n3

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Example 1 Use the comparison test to determine if the following series converges or diverges: ∞ X 2−1/n n3 n=1

Annette Pilkington Lecture 26 : Comparison Test √ We have 21/n = n 2 > 1 for n ≥ 1. Therefore 2−1/n = √1 < 1 for n ≥ 1. I n 2 2−1/n 1 I Therefore n3 < n3 for n > 1. P∞ 1 I Since n=1 n3 is a p-series with p > 1, it converges. P∞ 1 I Comparing the above series with n=1 n3 , we can conclude that P∞ 2−1/n P∞ 2−1/n P∞ 1 n=1 n3 also converges and n=1 n3 ≤ n=1 n3

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Example 1 Use the comparison test to determine if the following series converges or diverges: ∞ X 2−1/n n3 n=1

2−1/n I First we check that an > 0 –> true since n3 > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test 2−1/n 1 I Therefore n3 < n3 for n > 1. P∞ 1 I Since n=1 n3 is a p-series with p > 1, it converges. P∞ 1 I Comparing the above series with n=1 n3 , we can conclude that P∞ 2−1/n P∞ 2−1/n P∞ 1 n=1 n3 also converges and n=1 n3 ≤ n=1 n3

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Example 1 Use the comparison test to determine if the following series converges or diverges: ∞ X 2−1/n n3 n=1

2−1/n I First we check that an > 0 –> true since n3 > 0 for n ≥ 1. √ We have 21/n = n 2 > 1 for n ≥ 1. Therefore 2−1/n = √1 < 1 for n ≥ 1. I n 2

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Since n=1 n3 is a p-series with p > 1, it converges. P∞ 1 I Comparing the above series with n=1 n3 , we can conclude that P∞ 2−1/n P∞ 2−1/n P∞ 1 n=1 n3 also converges and n=1 n3 ≤ n=1 n3

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example 1

Example 1 Use the comparison test to determine if the following series converges or diverges: ∞ X 2−1/n n3 n=1

2−1/n I First we check that an > 0 –> true since n3 > 0 for n ≥ 1. √ We have 21/n = n 2 > 1 for n ≥ 1. Therefore 2−1/n = √1 < 1 for n ≥ 1. I n 2 2−1/n 1 I Therefore n3 < n3 for n > 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Comparing the above series with n=1 n3 , we can conclude that P∞ 2−1/n P∞ 2−1/n P∞ 1 n=1 n3 also converges and n=1 n3 ≤ n=1 n3

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Example 1 Use the comparison test to determine if the following series converges or diverges: ∞ X 2−1/n n3 n=1

2−1/n I First we check that an > 0 –> true since n3 > 0 for n ≥ 1. √ We have 21/n = n 2 > 1 for n ≥ 1. Therefore 2−1/n = √1 < 1 for n ≥ 1. I n 2 2−1/n 1 I Therefore n3 < n3 for n > 1. P∞ 1 I Since n=1 n3 is a p-series with p > 1, it converges. P∞ 1 I Comparing the above series with n=1 n3 , we can conclude that P∞ 2−1/n P∞ 2−1/n P∞ 1 n=1 n3 also converges and n=1 n3 ≤ n=1 n3

Annette Pilkington Lecture 26 : Comparison Test 21/n I First we check that an > 0 –> true since n > 0 for n ≥ 1. √ 1/n n I We have 2 = 2 > 1 for n ≥ 1. 21/n 1 I Therefore n > n for n > 1. P∞ 1 I Since n=1 n is a p-series with p = 1 (a.k.a. the harmonic series), it diverges. P∞ 21/n I Therefore, by comparison, we can conclude that n=1 n also diverges.

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Example 2 Use the comparison test to determine if the following series converges or diverges: ∞ X 21/n n n=1

Annette Pilkington Lecture 26 : Comparison Test √ 1/n n I We have 2 = 2 > 1 for n ≥ 1. 21/n 1 I Therefore n > n for n > 1. P∞ 1 I Since n=1 n is a p-series with p = 1 (a.k.a. the harmonic series), it diverges. P∞ 21/n I Therefore, by comparison, we can conclude that n=1 n also diverges.

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Example 2 Use the comparison test to determine if the following series converges or diverges: ∞ X 21/n n n=1

21/n I First we check that an > 0 –> true since n > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test 21/n 1 I Therefore n > n for n > 1. P∞ 1 I Since n=1 n is a p-series with p = 1 (a.k.a. the harmonic series), it diverges. P∞ 21/n I Therefore, by comparison, we can conclude that n=1 n also diverges.

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Example 2 Use the comparison test to determine if the following series converges or diverges: ∞ X 21/n n n=1

21/n I First we check that an > 0 –> true since n > 0 for n ≥ 1. √ 1/n n I We have 2 = 2 > 1 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Since n=1 n is a p-series with p = 1 (a.k.a. the harmonic series), it diverges. P∞ 21/n I Therefore, by comparison, we can conclude that n=1 n also diverges.

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Example 2 Use the comparison test to determine if the following series converges or diverges: ∞ X 21/n n n=1

21/n I First we check that an > 0 –> true since n > 0 for n ≥ 1. √ 1/n n I We have 2 = 2 > 1 for n ≥ 1. 21/n 1 I Therefore n > n for n > 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 21/n I Therefore, by comparison, we can conclude that n=1 n also diverges.

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Example 2 Use the comparison test to determine if the following series converges or diverges: ∞ X 21/n n n=1

21/n I First we check that an > 0 –> true since n > 0 for n ≥ 1. √ 1/n n I We have 2 = 2 > 1 for n ≥ 1. 21/n 1 I Therefore n > n for n > 1. P∞ 1 I Since n=1 n is a p-series with p = 1 (a.k.a. the harmonic series), it diverges.

Example 2 Use the comparison test to determine if the following series converges or diverges: ∞ X 21/n n n=1

Annette Pilkington Lecture 26 : Comparison Test 1 I First we check that an > 0 –> true since n2+1 > 0 for n ≥ 1. 2 2 I We have n + 1 > n for n ≥ 1. 1 1 I Therefore n2+1 < n2 for n > 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n2+1 also converges P∞ 1 P∞ 1 and n=1 n2+1 ≤ n=1 n2 .

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Example 3 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n2 + 1 n=1

Annette Pilkington Lecture 26 : Comparison Test 2 2 I We have n + 1 > n for n ≥ 1. 1 1 I Therefore n2+1 < n2 for n > 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n2+1 also converges P∞ 1 P∞ 1 and n=1 n2+1 ≤ n=1 n2 .

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Example 3 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n2 + 1 n=1

1 I First we check that an > 0 –> true since n2+1 > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test 1 1 I Therefore n2+1 < n2 for n > 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n2+1 also converges P∞ 1 P∞ 1 and n=1 n2+1 ≤ n=1 n2 .

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Example 3 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n2 + 1 n=1

1 I First we check that an > 0 –> true since n2+1 > 0 for n ≥ 1. 2 2 I We have n + 1 > n for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n2+1 also converges P∞ 1 P∞ 1 and n=1 n2+1 ≤ n=1 n2 .

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Example 3 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n2 + 1 n=1

1 I First we check that an > 0 –> true since n2+1 > 0 for n ≥ 1. 2 2 I We have n + 1 > n for n ≥ 1. 1 1 I Therefore n2+1 < n2 for n > 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Therefore, by comparison, we can conclude that n=1 n2+1 also converges P∞ 1 P∞ 1 and n=1 n2+1 ≤ n=1 n2 .

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Example 3 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n2 + 1 n=1

1 I First we check that an > 0 –> true since n2+1 > 0 for n ≥ 1. 2 2 I We have n + 1 > n for n ≥ 1. 1 1 I Therefore n2+1 < n2 for n > 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges.

Example 3 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n2 + 1 n=1

1 I First we check that an > 0 –> true since n2+1 > 0 for n ≥ 1. 2 2 I We have n + 1 > n for n ≥ 1. 1 1 I Therefore n2+1 < n2 for n > 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n2+1 also converges P∞ 1 P∞ 1 and n=1 n2+1 ≤ n=1 n2 .

Annette Pilkington Lecture 26 : Comparison Test n−2 1 I First we check that an > 0 –> true since 2n = n22n > 0 for n ≥ 1. 1 1 I We have n22n < n2 for n ≥ 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n22n also converges P∞ 1 P∞ 1 and n=1 n22n ≤ n=1 n2 .

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Example 4 Use the comparison test to determine if the following series converges or diverges: ∞ X n−2 2n n=1

Annette Pilkington Lecture 26 : Comparison Test 1 1 I We have n22n < n2 for n ≥ 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n22n also converges P∞ 1 P∞ 1 and n=1 n22n ≤ n=1 n2 .

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Example 4 Use the comparison test to determine if the following series converges or diverges: ∞ X n−2 2n n=1

n−2 1 I First we check that an > 0 –> true since 2n = n22n > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n22n also converges P∞ 1 P∞ 1 and n=1 n22n ≤ n=1 n2 .

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Example 4 Use the comparison test to determine if the following series converges or diverges: ∞ X n−2 2n n=1

n−2 1 I First we check that an > 0 –> true since 2n = n22n > 0 for n ≥ 1. 1 1 I We have n22n < n2 for n ≥ 1.

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Example 4 Use the comparison test to determine if the following series converges or diverges: ∞ X n−2 2n n=1

n−2 1 I First we check that an > 0 –> true since 2n = n22n > 0 for n ≥ 1. 1 1 I We have n22n < n2 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I Therefore, by comparison, we can conclude that n=1 n22n also converges P∞ 1 P∞ 1 and n=1 n22n ≤ n=1 n2 .

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Example 4 Use the comparison test to determine if the following series converges or diverges: ∞ X n−2 2n n=1

n−2 1 I First we check that an > 0 –> true since 2n = n22n > 0 for n ≥ 1. 1 1 I We have n22n < n2 for n ≥ 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges.

Example 4 Use the comparison test to determine if the following series converges or diverges: ∞ X n−2 2n n=1

n−2 1 I First we check that an > 0 –> true since 2n = n22n > 0 for n ≥ 1. 1 1 I We have n22n < n2 for n ≥ 1. P∞ 1 I Since n=1 n2 is a p-series with p = 2, it converges. P∞ 1 I Therefore, by comparison, we can conclude that n=1 n22n also converges P∞ 1 P∞ 1 and n=1 n22n ≤ n=1 n2 .

Annette Pilkington Lecture 26 : Comparison Test ln n 1 I First we check that an > 0 –> true since n > n > 0 for n ≥ e. Note that this allows us to use the test since a ﬁnite number of terms have no bearing on convergence or divergence. ln n 1 I We have n > n for n > 3. P∞ 1 P∞ ln n I Since n=1 n diverges, we can conclude that n=1 n also diverges.

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Example 5 Use the comparison test to determine if the following series converges or diverges: ∞ X ln n n n=1

Annette Pilkington Lecture 26 : Comparison Test ln n 1 I We have n > n for n > 3. P∞ 1 P∞ ln n I Since n=1 n diverges, we can conclude that n=1 n also diverges.

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Example 5 Use the comparison test to determine if the following series converges or diverges: ∞ X ln n n n=1

ln n 1 I First we check that an > 0 –> true since n > n > 0 for n ≥ e. Note that this allows us to use the test since a ﬁnite number of terms have no bearing on convergence or divergence.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 P∞ ln n I Since n=1 n diverges, we can conclude that n=1 n also diverges.

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Example 5 Use the comparison test to determine if the following series converges or diverges: ∞ X ln n n n=1

ln n 1 I First we check that an > 0 –> true since n > n > 0 for n ≥ e. Note that this allows us to use the test since a ﬁnite number of terms have no bearing on convergence or divergence. ln n 1 I We have n > n for n > 3.

Example 5 Use the comparison test to determine if the following series converges or diverges: ∞ X ln n n n=1

Annette Pilkington Lecture 26 : Comparison Test 1 I First we check that an > 0 –> true since n! > 0 for n ≥ 1. n−1 I We have n! = n(n − 1)(n − 2) ···· 2 · 1 > 2 · 2 · 2 ····· 2 · 1 = 2 . 1 1 Therefore n! < 2n−1 . P∞ 1 P∞ 1 I Since n=1 2n−1 converges, we can conclude that n=1 n! also converges.

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Example 6 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n! n=1

Annette Pilkington Lecture 26 : Comparison Test n−1 I We have n! = n(n − 1)(n − 2) ···· 2 · 1 > 2 · 2 · 2 ····· 2 · 1 = 2 . 1 1 Therefore n! < 2n−1 . P∞ 1 P∞ 1 I Since n=1 2n−1 converges, we can conclude that n=1 n! also converges.

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Example 6 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n! n=1

1 I First we check that an > 0 –> true since n! > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 P∞ 1 I Since n=1 2n−1 converges, we can conclude that n=1 n! also converges.

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Example 6 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n! n=1

1 I First we check that an > 0 –> true since n! > 0 for n ≥ 1. n−1 I We have n! = n(n − 1)(n − 2) ···· 2 · 1 > 2 · 2 · 2 ····· 2 · 1 = 2 . 1 1 Therefore n! < 2n−1 .

Example 6 Use the comparison test to determine if the following series converges or diverges: ∞ X 1 n! n=1

1 I First we check that an > 0 –> true since n! > 0 for n ≥ 1. n−1 I We have n! = n(n − 1)(n − 2) ···· 2 · 1 > 2 · 2 · 2 ····· 2 · 1 = 2 . 1 1 Therefore n! < 2n−1 . P∞ 1 P∞ 1 I Since n=1 2n−1 converges, we can conclude that n=1 n! also converges.

Annette Pilkington Lecture 26 : Comparison Test 1 I First we check that an > 0 –> true since an = n2−1 > 0 for n ≥ 2. (after we study absolute convergence, we see how to get around this restriction.) P∞ 1 I We will compare this series to n=2 n2 which converges, since it is a 1 p-series with p = 2. bn = n2 . 2 2 an 1/(n −1) n 1 limn→∞ = limn→∞ = limn→∞ = limn→∞ = 1 I bn 1/n2 n2−1 1−(1/n2) I Since c = 1 > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I We will compare this series to n=2 n2 which converges, since it is a 1 p-series with p = 2. bn = n2 . 2 2 an 1/(n −1) n 1 limn→∞ = limn→∞ = limn→∞ = limn→∞ = 1 I bn 1/n2 n2−1 1−(1/n2) I Since c = 1 > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Limit Comparison Test P P Limit Comparison Test Suppose that an and bn are series with positive terms. If a lim n = c n→∞ bn where c is a ﬁnite number and c > 0, then either both series converge or both diverge. (Note c 6= 0 or ∞.) Example Test the following series for convergence using the Limit Comparison test: ∞ X 1 n2 − 1 n=2 (Note that our previous comparison test is diﬃcult to apply in this and most of the examples below.) 1 I First we check that an > 0 –> true since an = n2−1 > 0 for n ≥ 2. (after we study absolute convergence, we see how to get around this restriction.)

Annette Pilkington Lecture 26 : Comparison Test 2 2 an 1/(n −1) n 1 limn→∞ = limn→∞ = limn→∞ = limn→∞ = 1 I bn 1/n2 n2−1 1−(1/n2) I Since c = 1 > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test I Since c = 1 > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Limit Comparison Test P P Limit Comparison Test Suppose that an and bn are series with positive terms. If a lim n = c n→∞ bn where c is a ﬁnite number and c > 0, then either both series converge or both diverge. (Note c 6= 0 or ∞.) Example Test the following series for convergence using the Limit Comparison test: ∞ X 1 n2 − 1 n=2 (Note that our previous comparison test is diﬃcult to apply in this and most of the examples below.) 1 I First we check that an > 0 –> true since an = n2−1 > 0 for n ≥ 2. (after we study absolute convergence, we see how to get around this restriction.) P∞ 1 I We will compare this series to n=2 n2 which converges, since it is a 1 p-series with p = 2. bn = n2 . 2 2 an 1/(n −1) n 1 limn→∞ = limn→∞ = limn→∞ = limn→∞ = 1 I bn 1/n2 n2−1 1−(1/n2) I Since c = 1 > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test n2+2n+1 I First we check that an > 0 –> true since an = n4+n2+2n+1 > 0 for n ≥ 1. I For a rational function, the rule of thumb is to compare the series to the P np series nq , where p is the degree of the numerator and q is the degree of the denominator. P∞ n2 P∞ 1 I We will compare this series to n=1 n4 = n=1 n2 which converges, since 1 it is a p-series with p = 2. bn = n2 . 2 4 3 2 an n +2n+1 2 n +2n +n limn→∞ = limn→∞( )/(1/n ) = limn→∞ = I bn n4+n2+2n+1 n4+n2+2n+1 1+2/n+1/n2 limn→∞ 1+1/n2+2/n3+1/n4 = 1.

I Since c = 1 > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example P P Limit Comparison Test Suppose that an and bn are series with an positive terms. If limn→∞ = c where c is a ﬁnite number and c > 0, then bn either both series converge or both diverge. (Note c 6= 0 or ∞.) Example Test the following series for convergence using the Limit Comparison test: ∞ X n2 + 2n + 1 n4 + n2 + 2n + 1 n=1

Annette Pilkington Lecture 26 : Comparison Test I For a rational function, the rule of thumb is to compare the series to the P np series nq , where p is the degree of the numerator and q is the degree of the denominator. P∞ n2 P∞ 1 I We will compare this series to n=1 n4 = n=1 n2 which converges, since 1 it is a p-series with p = 2. bn = n2 . 2 4 3 2 an n +2n+1 2 n +2n +n limn→∞ = limn→∞( )/(1/n ) = limn→∞ = I bn n4+n2+2n+1 n4+n2+2n+1 1+2/n+1/n2 limn→∞ 1+1/n2+2/n3+1/n4 = 1.

I Since c = 1 > 0, we can conclude that both series converge.

n2+2n+1 I First we check that an > 0 –> true since an = n4+n2+2n+1 > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test P∞ n2 P∞ 1 I We will compare this series to n=1 n4 = n=1 n2 which converges, since 1 it is a p-series with p = 2. bn = n2 . 2 4 3 2 an n +2n+1 2 n +2n +n limn→∞ = limn→∞( )/(1/n ) = limn→∞ = I bn n4+n2+2n+1 n4+n2+2n+1 1+2/n+1/n2 limn→∞ 1+1/n2+2/n3+1/n4 = 1.

I Since c = 1 > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test 2 4 3 2 an n +2n+1 2 n +2n +n limn→∞ = limn→∞( )/(1/n ) = limn→∞ = I bn n4+n2+2n+1 n4+n2+2n+1 1+2/n+1/n2 limn→∞ 1+1/n2+2/n3+1/n4 = 1.

I Since c = 1 > 0, we can conclude that both series converge.

n2+2n+1 I First we check that an > 0 –> true since an = n4+n2+2n+1 > 0 for n ≥ 1. I For a rational function, the rule of thumb is to compare the series to the P np series nq , where p is the degree of the numerator and q is the degree of the denominator. P∞ n2 P∞ 1 I We will compare this series to n=1 n4 = n=1 n2 which converges, since 1 it is a p-series with p = 2. bn = n2 .

Annette Pilkington Lecture 26 : Comparison Test I Since c = 1 > 0, we can conclude that both series converge.

I Since c = 1 > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test 2n+1 I First we check that an > 0 –> true since an = √ > 0 for n ≥ 1. n3+1 ∞ ∞ ∞ We will compare this series to P √n = P n = P √1 which I n=1 n3 n=1 n3/2 n=1 n 1 diverges, since it is a p-series with p = 1/2. b = √ . n n

. √ 3/2 √ an 2n+1 +2n + n I limn→∞ = limn→∞( √ ) (1/ n) = limn→∞ √ = bn n3+1 n3+1 3/2 √ ‹ 3/2 (2n + n) n (2+1/n) (2+1/n) limn→∞ √ ‹ = limn→∞ √ = limn→∞ √ = 2. n3+1 n3/2 (n3+1)/n3 (1+1/n3)

I Since c = 2 > 0, we can conclude that both series diverge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 2n + 1 √ 3 n=1 n + 1

Annette Pilkington Lecture 26 : Comparison Test ∞ ∞ ∞ We will compare this series to P √n = P n = P √1 which I n=1 n3 n=1 n3/2 n=1 n 1 diverges, since it is a p-series with p = 1/2. b = √ . n n

I Since c = 2 > 0, we can conclude that both series diverge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 2n + 1 √ 3 n=1 n + 1

2n+1 I First we check that an > 0 –> true since an = √ > 0 for n ≥ 1. n3+1

Annette Pilkington Lecture 26 : Comparison Test . √ 3/2 √ an 2n+1 +2n + n I limn→∞ = limn→∞( √ ) (1/ n) = limn→∞ √ = bn n3+1 n3+1 3/2 √ ‹ 3/2 (2n + n) n (2+1/n) (2+1/n) limn→∞ √ ‹ = limn→∞ √ = limn→∞ √ = 2. n3+1 n3/2 (n3+1)/n3 (1+1/n3)

I Since c = 2 > 0, we can conclude that both series diverge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 2n + 1 √ 3 n=1 n + 1

2n+1 I First we check that an > 0 –> true since an = √ > 0 for n ≥ 1. n3+1 ∞ ∞ ∞ We will compare this series to P √n = P n = P √1 which I n=1 n3 n=1 n3/2 n=1 n 1 diverges, since it is a p-series with p = 1/2. b = √ . n n

Annette Pilkington Lecture 26 : Comparison Test I Since c = 2 > 0, we can conclude that both series diverge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 2n + 1 √ 3 n=1 n + 1

Annette Pilkington Lecture 26 : Comparison Test I Since c = 2 > 0, we can conclude that both series diverge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 2n + 1 √ 3 n=1 n + 1

I Since c = 2 > 0, we can conclude that both series diverge.

Annette Pilkington Lecture 26 : Comparison Test e I First we check that an > 0 –> true since an = 2n−1 > 0 for n ≥ 1. P∞ 1 I We will compare this series to n=1 2n which converges, since it is a 1 geometric series with r = 1/2 < 1. b = . n 2n . an e n e limn→∞ = limn→∞( n ) (1/2 ) = limn→∞ n = e. I bn 2 −1 1−1/2

I Since c = e > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X e 2n − 1 n=1

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I We will compare this series to n=1 2n which converges, since it is a 1 geometric series with r = 1/2 < 1. b = . n 2n . an e n e limn→∞ = limn→∞( n ) (1/2 ) = limn→∞ n = e. I bn 2 −1 1−1/2

I Since c = e > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X e 2n − 1 n=1

e I First we check that an > 0 –> true since an = 2n−1 > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test . an e n e limn→∞ = limn→∞( n ) (1/2 ) = limn→∞ n = e. I bn 2 −1 1−1/2

I Since c = e > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X e 2n − 1 n=1

e I First we check that an > 0 –> true since an = 2n−1 > 0 for n ≥ 1. P∞ 1 I We will compare this series to n=1 2n which converges, since it is a 1 geometric series with r = 1/2 < 1. b = . n 2n

Annette Pilkington Lecture 26 : Comparison Test I Since c = e > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X e 2n − 1 n=1

e I First we check that an > 0 –> true since an = 2n−1 > 0 for n ≥ 1. P∞ 1 I We will compare this series to n=1 2n which converges, since it is a 1 geometric series with r = 1/2 < 1. b = . n 2n . an e n e limn→∞ = limn→∞( n ) (1/2 ) = limn→∞ n = e. I bn 2 −1 1−1/2

Example Test the following series for convergence using the Limit Comparison test: ∞ X e 2n − 1 n=1

I Since c = e > 0, we can conclude that both series converge.

Annette Pilkington Lecture 26 : Comparison Test 21/n I First we check that an > 0 –> true since an = n2 > 0 for n ≥ 1. P∞ 1 I We will compare this series to n=1 n2 which converges, since it is a 1 p-series with p = 2 > 1. b = . n n2

1/n . ln 2 an 2 2 1/n limn→∞ = limn→∞( ) (1/n ) = limn→∞ 2 = limn→∞ e n = 1. I bn n2

I Since c = 1 > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 21/n n2 n=1

Annette Pilkington Lecture 26 : Comparison Test P∞ 1 I We will compare this series to n=1 n2 which converges, since it is a 1 p-series with p = 2 > 1. b = . n n2

1/n . ln 2 an 2 2 1/n limn→∞ = limn→∞( ) (1/n ) = limn→∞ 2 = limn→∞ e n = 1. I bn n2

I Since c = 1 > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 21/n n2 n=1

21/n I First we check that an > 0 –> true since an = n2 > 0 for n ≥ 1.

Annette Pilkington Lecture 26 : Comparison Test 1/n . ln 2 an 2 2 1/n limn→∞ = limn→∞( ) (1/n ) = limn→∞ 2 = limn→∞ e n = 1. I bn n2

I Since c = 1 > 0, we can conclude that both series converge.

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example Example

Example Test the following series for convergence using the Limit Comparison test: ∞ X 21/n n2 n=1

21/n I First we check that an > 0 –> true since an = n2 > 0 for n ≥ 1. P∞ 1 I We will compare this series to n=1 n2 which converges, since it is a 1 p-series with p = 2 > 1. b = . n n2