24 the harmonic series and the integral test x

The Harmonic Series and the Integral Test

1. The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero.

2. Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞

3. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞

4. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: This means that for the sequence of partial sums, lim s n = lim (a 1 + a 2 + … + a n ) = L converges. The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. n  ∞ If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞

5. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. n  ∞ lim s n-1 = lim (a 1 + a 2 +..+ a n-1 ) n  ∞ n  ∞ If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞ On the other hand, This means that for the sequence of partial sums, lim s n = lim (a 1 + a 2 + … + a n ) = L converges.

6. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. n  ∞ lim s n-1 = lim (a 1 + a 2 +..+ a n-1 ) = L. n  ∞ n  ∞ If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞ On the other hand, This means that for the sequence of partial sums, lim s n = lim (a 1 + a 2 + … + a n ) = L converges.

7. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. n  ∞ lim s n-1 = lim (a 1 + a 2 +..+ a n-1 ) = L. n  ∞ n  ∞ Hence lim s n – s n-1 n  ∞ If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞ On the other hand, This means that for the sequence of partial sums, lim s n = lim (a 1 + a 2 + … + a n ) = L converges.

8. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. n  ∞ lim s n-1 = lim (a 1 + a 2 +..+ a n-1 ) = L. n  ∞ n  ∞ Hence lim s n – s n-1 = lim a n n  ∞ n  ∞ If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞ On the other hand, This means that for the sequence of partial sums, lim s n = lim (a 1 + a 2 + … + a n ) = L converges.

9. Proof: Let = a 1 + a 2 .. = L be a convergent series. Σ i=1 ∞ a i Theorem: The Harmonic Series and the Integral Test If we add infinitely many terms and obtain a finite sum, it must be the case that the terms get smaller and smaller and go to zero. n  ∞ lim s n-1 = lim (a 1 + a 2 +..+ a n-1 ) = L. n  ∞ n  ∞ Hence lim s n – s n-1 = lim a n = L – L = 0. n  ∞ n  ∞ If = a 1 + a 2 + a 3 + … = L is a Σ i=1 ∞ a i convergent series, then lim a n = 0. n  ∞ On the other hand, This means that for the sequence of partial sums, lim s n = lim (a 1 + a 2 + … + a n ) = L converges.

10. The Harmonic Series and the Integral Test However the fact that lim a n  does not guarantee that their sum converges.

11. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , ..

12. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum

13. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1

14. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2

15. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 + + +

16. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 + + + + + + +

17. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞

18. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { }

19. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown:

20. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1

21. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + +

22. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + > 9 10

23. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + 1 11 + 1 100 ... + > 9 10

24. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + 1 11 + 1 100 ... + > 9 10 > 90 100 = 9 10

25. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + 1 11 + 1 100 ... + 1 101 + 1 1000 ... + > 9 10 > 90 100 = 9 10

26. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + 1 11 + 1 100 ... + 1 101 + 1 1000 ... + > 9 10 > 90 100 = 9 10 > 1000 = 10 900 9

27. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + 1 11 + 1 100 ... + 1 101 + 1 1000 ... + + … > 9 10 > 90 100 = 9 10 > 1000 = 10 900 9 = ∞

28. Example A. The sequence 1, The Harmonic Series and the Integral Test However the fact that lim a n  0 does not guarantee that their sum converges. Here is an example. 1 2 , 1 2 , 1 3 , 1 3 , 1 3 , 1 4 , 1 4 , 1 4 , 1 4 ,  0, 1 5 , .. but their sum 1+ 1 2 + 1 2 1 3 1 3 1 3 1 4 1 4 1 4 1 4 1 5 .. + + + + + + + + = ∞ An important sequence that goes to 0 but sums to ∞ is the harmonic sequence: {1/n} = 1 2 , 1 3 , 1 4 , .. 1, { } To see that they sum to ∞ , sum in blocks as shown: 1 2 1 3 1 10 ... 1 + + + + 1 11 + 1 100 ... + 1 101 + 1 1000 ... + + … > 9 10 > 90 100 = 9 10 > 1000 = 10 900 9 = ∞ Hence the harmonic series diverges.

29. The Harmonic Series and the Integral Test The divergent harmonic series Σ n =1 ∞ 1/n = 1 2 1 3 ... 1 + + + = ∞ Area = ∞ y = 1/x

30. The Harmonic Series and the Integral Test The divergent harmonic series Σ n =1 ∞ 1/n = 1 2 1 3 ... 1 + + + = ∞ ∫ 1 x 1 dx = ∞ ∞ is the discrete version of the improper integral Area = ∞ Area = ∞ y = 1/x y = 1/x

31. The Harmonic Series and the Integral Test The divergent harmonic series Σ n =1 ∞ 1/n = 1 2 1 3 ... 1 + + + = ∞ ∫ 1 x 1 dx = ∞ The series 1/n p are called p-series and they correspond to the p-functions 1/ x p where 1 ≤ x < ∞ . Σ n =1 ∞ ∞ is the discrete version of the improper integral Area = ∞ Area = ∞ y = 1/x y = 1/x

32. The Harmonic Series and the Integral Test The divergent harmonic series the convergent and divergent p-series Σ n =1 ∞ 1/n = 1 2 1 3 ... 1 + + + = ∞ ∫ 1 x 1 dx = ∞ The series 1/n p are called p-series and they correspond to the p-functions 1/ x p where 1 ≤ x < ∞ . Just as the function 1/x serves as the boundary between the convergent and divergent 1/ x p , the harmonic series 1/n serves as the boundary of Σ n =1 ∞ Σ n =1 ∞ Σ 1/ n p . n =1 ∞ ∞ is the discrete version of the improper integral Area = ∞ Area = ∞ y = 1/x y = 1/x

33. The Harmonic Series and the Integral Test The divergent harmonic series the convergent and divergent p-series Σ n =1 ∞ 1/n = 1 2 1 3 ... 1 + + + = ∞ ∫ 1 x 1 dx = ∞ The series 1/n p are called p-series and they correspond to the p-functions 1/ x p where 1 ≤ x < ∞ . Just as the function 1/x serves as the boundary between the convergent and divergent 1/ x p , the harmonic series 1/n serves as the boundary of Σ n =1 ∞ Σ n =1 ∞ The following theorem establishes the connection between the discrete and the continuous cases. Σ 1/ n p . n =1 ∞ ∞ is the discrete version of the improper integral Area = ∞ Area = ∞ y = 1/x y = 1/x

34. Σ n =1 Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ The Integral Test ∫ f(x) dx converge or both diverge. 1 a n and both ∞

35. Σ n =1 Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ The Integral Test ∫ f(x) dx converge or both diverge. 1 a n and both ∞ In short, the series and the integrals behave the same. We leave the proof to the end of the section.

36. Σ n =1 Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ The Integral Test ∫ f(x) dx converge or both diverge. 1 a n and both ∞ In short, the series and the integrals behave the same. We leave the proof to the end of the section. Note that f(x) has to be a decreasing function from some point onward, i.e. f(x) can’t be a function that oscillates forever .

37. Σ n =1 Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ The Integral Test ∫ f(x) dx converge or both diverge. 1 a n and both ∞ In short, the series and the integrals behave the same. We leave the proof to the end of the section. Note that f(x) has to be a decreasing function from some point onward, i.e. f(x) can’t be a function that oscillates forever. For example, for all n, let a n = 0 = f(n) where f(x ) = sin 2 ( π x),

38. Σ n =1 Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ The Integral Test ∫ f(x) dx converge or both diverge. 1 a n and both ∞ In short, the series and the integrals behave the same. We leave the proof to the end of the section. Note that f(x) has to be a decreasing function from some point onward, i.e. f(x) can’t be a function that oscillates forever. For example, for all n, let a n = 0 = f(n) where f(x ) = sin 2 ( π x), f(x) = ( sin( π x )) 2 (1, 0) (2, 0) (3, 0)

39. Σ n =1 Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ The Integral Test ∫ f(x) dx converge or both diverge. 1 a n and both ∞ In short, the series and the integrals behave the same. We leave the proof to the end of the section. Note that f(x) has to be a decreasing function from some point onward, i.e. f(x) can’t be a function that oscillates forever. For example, for all n, let a n = 0 = f(n) where f(x ) = sin 2 ( π x), f(x) = ( sin( π x )) 2 (1, 0) (2, 0) (3, 0) Σ n =1 ∞ ∫ f(x) dx diverges. 1 a n = 0 = ∞ Σ n = 1 ∞ sin 2 ( π x ) converges, but

40. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1

41. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1 Proof : The functions y = 1/ x p are positive decreasing functions hence the integral test applies.

42. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1 Proof : The functions y = 1/ x p are positive decreasing functions hence the integral test applies. ∫ 1 x p 1 ∞ dx converges if and only if p > 1 , Since Σ n=1 converges if and only if p > 1. ∞ n p 1 so

43. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1 Proof : The functions y = 1/ x p are positive decreasing functions hence the integral test applies. n 1.01 1 converges and that Σ n=1 ∞ So n .99 1 Σ n=1 ∞ diverges. ∫ 1 x p 1 ∞ dx converges if and only if p > 1 , Since Σ n=1 converges if and only if p > 1. ∞ n p 1 so

44. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1 Proof : The functions y = 1/ x p are positive decreasing functions hence the integral test applies. n 1.01 1 converges and that Σ n=1 ∞ So n .99 1 Σ n=1 ∞ diverges. ∫ 1 x p 1 ∞ dx converges if and only if p > 1 , Since Σ n=1 converges if and only if p > 1. ∞ n p 1 so Recall the floor and ceiling theorems for integrals:

45. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1 Proof : The functions y = 1/ x p are positive decreasing functions hence the integral test applies. n 1.01 1 converges and that Σ n=1 ∞ So n .99 1 Σ n=1 ∞ diverges. ∫ 1 x p 1 ∞ dx converges if and only if p > 1 , Since Σ n=1 converges if and only if p > 1. ∞ n p 1 so Recall the floor and ceiling theorems for integrals: I. if f(x) ≥ g(x) ≥ 0 and g(x ) dx = ∞, then f(x ) = ∞. ∫ a b ∫ a b

46. Σ n=1 Theorem ( p-series) The Harmonic Series and the Integral Test converges if and only if p > 1. ∞ n p 1 Proof : The functions y = 1/ x p are positive decreasing functions hence the integral test applies. n 1.01 1 converges and that Σ n=1 ∞ So n .99 1 Σ n=1 ∞ diverges. ∫ 1 x p 1 ∞ dx converges if and only if p > 1 , Since Σ n=1 converges if and only if p > 1. ∞ n p 1 so Recall the floor and ceiling theorems for integrals: I. if f(x) ≥ g(x) ≥ 0 and g(x ) dx = ∞, then f(x ) = ∞. ∫ a b ∫ a b II. if f(x) ≥ g(x) ≥ 0 and f(x ) dx < ∞, then g (x ) < ∞. ∫ a b ∫ a b

47. The Harmonic Series and the Integral Test By the same logic we have their discrete versions.

48. The Harmonic Series and the Integral Test By the same logic we have their discrete versions. The Floor Theorem

49. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ By the same logic we have their discrete versions. The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0.

50. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ Example B. Does converge or diverge? By the same logic we have their discrete versions. Σ n =2 ∞ Ln (n) 1 The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0.

51. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ Example B. Does converge or diverge? Ln (n) 1 > n . 1 For n > 1, n > Ln(n) (why?) By the same logic we have their discrete versions. Σ n =2 ∞ Ln (n) 1 so The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0.

52. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ Example B. Does converge or diverge? Ln (n) 1 > n . 1 For n > 1, n > Ln(n) (why?) Σ n =2 n 1 t herefore Σ n =2 Ln (n) 1 By the same logic we have their discrete versions. > = ∞ diverges. Σ n =2 ∞ Ln (n) 1 so The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =2 n 1 = ∞ Since

53. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ Example B. Does converge or diverge? Ln (n) 1 > n . 1 For n > 1, n > Ln(n) (why?) Σ n =2 n 1 t herefore Σ n =2 Ln (n) 1 By the same logic we have their discrete versions. > = ∞ diverges. Σ n =2 ∞ Ln (n) 1 so The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. No conclusion can be drawn about Σ a n if a n ≥ b n and Σ b n < ∞. Σ n =2 n 1 = ∞ Since

54. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ Example B. Does converge or diverge? Ln (n) 1 > n . 1 For n > 1, n > Ln(n) (why?) Σ n =2 n 1 t herefore Σ n =2 Ln (n) 1 By the same logic we have their discrete versions. > = ∞ diverges. Σ n =2 ∞ Ln (n) 1 so The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. No conclusion can be drawn about Σ a n if a n ≥ b n and Σ b n < ∞. For example, both 1/n and 1/n 3/2 > b n = 1/n 2 and Σ 1/n 2 = Σ b n < ∞ since p = 2 > 1, Σ n =2 n 1 = ∞ Since

55. The Harmonic Series and the Integral Test Suppose b n = ∞, then a n = ∞. Σ n =k ∞ Σ n =k ∞ Example B. Does converge or diverge? Ln (n) 1 > n . 1 For n > 1, n > Ln(n) (why?) Σ n =2 n 1 t herefore Σ n =2 Ln (n) 1 By the same logic we have their discrete versions. > = ∞ diverges. Σ n =2 ∞ Ln (n) 1 so The Floor Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. No conclusion can be drawn about Σ a n if a n ≥ b n and Σ b n < ∞. For example, both 1/n and 1/n 3/2 > b n = 1/n 2 and Σ 1/n 2 = Σ b n < ∞ since p = 2 > 1, h owever Σ 1/n = ∞ diverges but Σ 1/n 3/2 < ∞ converges, so in general no conclusion can be made about Σ a n . Σ n =2 n 1 = ∞ Since

56. The Harmonic Series and the Integral Test The Ceiling Theorem

57. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0.

58. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k Example C. Does converge or diverge? The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =1 n 2 + 4 2

59. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k Example C. Does converge or diverge? The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =1 n 2 + 4 2 Compare with n 2 + 4 2 n 2 2

60. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k Example C. Does converge or diverge? n 2 + 4 2 > n 2 2 , we have n 2 + 4 The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =1 n 2 + 4 2 Compare with n 2 + 4 2 n 2 2 2 Σ Σ n 2 2 > so that

61. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k Example C. Does converge or diverge? n 2 + 4 2 > n 2 2 , Σ n 2 2 we have = N converges since p = 2 > 1, n 2 + 4 The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =1 n 2 + 4 2 Compare with n 2 + 4 2 n 2 2 2 = 2 Σ n 2 1 Σ Σ n 2 2 > so that

62. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k Example C. Does converge or diverge? n 2 + 4 2 > n 2 2 , Σ n 2 2 we have = N converges since p = 2 > 1, n 2 + 4 The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =1 n 2 + 4 2 Compare with n 2 + 4 2 n 2 2 2 = 2 Σ n 2 1 Σ Σ n 2 2 > so that therefore N > n 2 + 4 2 Σ so the series converges.

63. The Harmonic Series and the Integral Test If a n converges , then b n converges. Σ n =k Σ n =k Example C. Does converge or diverge? n 2 + 4 2 > n 2 2 , Σ n 2 2 we have = N converges since p = 2 > 1, n 2 + 4 The Ceiling Theorem Let {a n } and { b n } be two sequences and a n ≥ b n ≥ 0. Σ n =1 n 2 + 4 2 Compare with n 2 + 4 2 n 2 2 2 No conclusion can be drawn about Σ b n if a n ≥ b n and Σ a n = ∞ for Σ b n may converge or it may diverge. = 2 Σ n 2 1 Σ Σ n 2 2 > so that therefore N > n 2 + 4 2 Σ so the series converges.

64. Σ n =1 “Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ ∫ f(x) dx converge or both diverge.” 1 a n and both ∞ Here is an intuitive proof of the Integral Test

65. Σ n =1 “Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ ∫ f(x) dx converge or both diverge.” 1 a n and both ∞ Here is an intuitive proof of the Integral Test ∫ f(x ) dx converges, 1 ∞ I. so if S ince ∫ f(x) dx > 1 ∞ a 2 + a 3 + … then Σ a n converges.

66. Σ n =1 “Let a n = f(n) where f(x) is a positive continuous eventually decreasing function defined for 1 ≤ x, then The Harmonic Series and the Integral Test ∞ ∫ f(x) dx converge or both diverge.” 1 a n and both ∞ Here is an intuitive proof of the Integral Test ∫ f(x ) dx converges, 1 ∞ I. so if S ince ∫ f(x) dx > 1 ∞ a 2 + a 3 + … then Σ a n converges. Let g(x) = f(x + 1) so g(x ) is f(x ) shifted left by 1, l I. and that g(0 ) = a 1 , g(1) = a 2 , g(2 ) = a 3 ,. . etc. g(x ) is decreasing hence it’s below the rectangles. So if Σ a n converges f(x ) dx converges. 1 ∫ then ∞