Sequences and Series - Amazon S3 · 2015-04-02 · Sequences and Series 1/14/14 8:48 PM ... Partial Sums and Convergence of Series As in Section 9.2, we define the partial sum, Sn,

1/14/14 8:48 PMSequences and Series

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9.3 CONVERGENCE OF SERIESWe now consider general series in which each term an is a number. The series can be written compactly using a ∑sign as follows:

For any values of a and x , the geometric series is such a series, with general term an = ax n −1.

Partial Sums and Convergence of SeriesAs in Section 9.2, we define the partial sum, Sn , of the first n terms of a series as

To investigate the convergence of the series, we consider the sequence of partial sumsS1, S2, S3, �, Sn , �.

If Sn has a limit as n→∞, then we define the sum of the series to be that limit.

The following example shows how a series leads to sequence of partial sums and how we use them to determineconvergence.

If the sequence Sn of partial sums converges to S , so , then we say the series converges and that

its sum is S . We write . If does not exist, we say that the series diverges.

Investigate the convergence of the series with an = 1/ (n (n +1) ):Example 1

In order to determine whether the series converges, we first find the partial sums:

It appears that Sn = n / (n +1) for each positive integer n . We check that this pattern continues byassuming that Sn = n / (n +1) for a given integer n , adding an +1, and simplifying

Solution



Visualizing SeriesWe can visualize the terms of the series in Example 1 as the heights of the bars in Figure 9.4. The partial sums ofthe series are illustrated by stacking the bars on top of each other in Figure 9.5.

Figure 9.4 Terms of the series with an = 1/ (n (n +1) )

Figure 9.5 Partial sums of the series with an = 1/ (n (n +1) )

Here are some properties that are useful in determining whether or not a series converges.

Thus the sequence of partial sums has formula Sn = n / (n +1), which converges to 1, so the series converges to 1. That is, we can say that

Theorem 9.2: Convergence Properties of Series1. If and converge and if k is a constant, then

• converges to .

• converges to .



For proofs of these properties, see Problems 39-42. As for improper integrals, the convergence of a series isdetermined by its behavior for large n . (See the “behaves like” principle.) From Property 2 we see that, if N is a

positive integer, then and either both converge or both diverge. Thus, if all we care about is the

convergence of a series, we can omit the limits and write ∑an .

Comparison of Series and IntegralsWe investigate the convergence of some series by comparison with an improper integral. The harmonic series is theinfinite series

Convergence of this sum would mean that the sequence of partial sums

tends to a limit as n→∞. Let's look at some values:S1 = 1, S10≈2.93, S100≈5.19, S1000≈7.49, S10000≈9.79.

The growth of these partial sums is slow, but they do in fact grow without bound, so the harmonic series diverges.This is justified in the following example and in Problem 46.

2. Changing a finite number of terms in a series does not change whether or not it converges, although it maychange the value of its sum if it does converge.

3. If or does not exist, then diverges.

4. If diverges, then diverges if k ≠0.

Does the series ∑ (1−e −n ) converge?Example 2

Since the terms in the series, an = 1−e −n tend to 1, not 0, as n→∞, the series diverges byProperty 3 of Theorem 9.2.

Solution

Show that the harmonic series 1+1/2+1/3+1/4+… diverges.Example 3

The idea is to approximate by a left-hand sum, where the terms 1, 1/2, 1/3, … are

heights of rectangles of base 1. In Figure 9.6, the sum of the areas of the 3 rectangles is larger thanthe area under the curve between x = 1 and x = 4, and the same kind of relationship holds for thefirst n rectangles. Thus, we have

Since ln (n +1) gets arbitrarily large as n→∞, so do the partial sums, Sn . Thus, the partial sumshave no limit, so the series diverges.

Solution



Notice that the harmonic series diverges, even though . Although Property 3 of Theorem 9.2guarantees ∑an diverges if , it is possible for ∑an to either converge or diverge if . When wehave , we must investigate the series further to determine whether it converges or diverges.

Figure 9.6 Comparing the harmonic series to

By comparison with the improper integral , show that the following series

converges:

Example 4

Since we want to show that converges, we want to show that the partial sums of this

series tend to a limit. We do this by showing that the sequence of partial sums increases and isbounded above, so Theorem 9.1 applies.Each successive partial sum is obtained from the previous one by adding one more term in theseries. Since all the terms are positive, the sequence of partial sums is increasing.

To show that the partial sums of are bounded, we consider the right-hand sum

represented by the area of the rectangles in Figure 9.7. We start at x = 1, since the area under thecurve is infinite for 0 ≤ x ≤ 1. The shaded rectangles in Figure 9.7 suggest that:

The area under the graph is finite, since

Solution



Notice that we have shown that the series in the Example 4 converges, but we have not found its sum. The integralgives us a bound on the partial sums, but it does not give us the limit of the partial sums. Euler proved theremarkable fact that the sum is π 2/6.The method of Examples 3 and 4 can be used to prove the following theorem. See Problem 45.

Suppose f (x ) is continuous. Then if f (x ) is positive and decreasing for all x beyond some point, say c , the integraltest can be used.The integral test allows us to analyze a family of series, the p -series, and see how convergence depends on theparameter p .

Figure 9.7 Comparing to

To get Sn , we add 1 to both sides, giving

Thus, the sequence of partial sums is bounded above by 2. Hence, by Theorem 9.1 the sequence ofpartial sums converges, so the series converges.

Theorem 9.3: The Integral TestSuppose an = f (n ), where f (x ) is decreasing and positive.

• If converges, then ∑an converges.

• If diverges, then ∑an diverges.

For what values of p does the series converge?Example 5

If p ≤ 0, the terms in the series an = 1/n p do not tend to 0 as n→∞. Thus the series diverges forp ≤ 0.

Solution



1.

2.

3.

4.

5.

6.

7.

We can summarize Example 5 as follows:

Exercises and Problems for Section 9.3Exercises

In Exercises 1-3, find the first five terms of the sequence of partial sums.

In Exercises 4-7, use the integral test to decide whether the series converges or diverges.

If p >0, we compare to the integral . In Example 3 of Section 7.6 we saw

that the integral converges if p >1 and diverges if p ≤ 1. By the integral test, we conclude that∑1/n p converges if p >1 and diverges if p ≤ 1.

The p -series converges if p >1 and diverges if p ≤ 1.

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8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

Use comparison with to show that converges to a number less than or equal to 1/2.

Use comparison with to show that converges to a number less than or

equal to π /2.

In Exercises 10-12, explain why the integral test cannot be used to decide if the series converges or diverges.

Problems

In Problems 13-32, does the series converge or diverge?










18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.












30.

31.

32.

33.

34.

(a)

(b)

35.

(a)

(b)

36.

(a)

(b)

(c)

37.

(a)

38.

Show that diverges.

Show that converges.

Find the partial sum, Sn , of .

Does the series in part a converge or diverge?

Show r ln n = n ln r for positive numbers n and r .

For what values r >0 does converge?

Consider the series .

Show that .

Use part a to find the partial sums S3, S10, and Sn .

Use part b to show that the sequence of partial sums Sn , and therefore the series, converges to 1.

Consider the series

Show that the partial sum of the first three nonzero terms S3 = ln (5/8).










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(b)

(c) 39.

40.

41.

42. 43. 44.

(a)

(b)

45.

(a) (b)

46.

(a) (b)

47.

48.

Show that the partial sum .

Use part b to show that the partial sums Sn , and therefore the series, converge to ln (1/2).Show that if ∑an and ∑bn converge and if k is a constant, then ∑ (an +bn ), ∑ (an −bn ), and ∑kan

converge.Let N be a positive integer. Show that if an = bn for n ≥N , then ∑an and ∑bn either both converge, or both

diverge.Show that if ∑an converges, then

[Hint: Consider limn→∞ (Sn −Sn −1), where Sn is the n th partial sum.]

Show that if ∑an diverges and k ≠0, then ∑kan diverges.The series ∑an converges. Explain, by looking at partial sums, why the series ∑ (an +1−an ) also converges.The series ∑an diverges. Give examples that show the series ∑ (an +1−an ) could converge or diverge.In this problem, you will justify the integral test. Suppose c ≥0 and f (x ) is a decreasing positive function,

defined for all numbers x ≥c , with f (n ) = an for all integers n ≥c .

Suppose that diverges. By considering rectangles above the graph of f , show that ∑an diverges.

Hint: See Example 3.

Suppose converges. By considering rectangles under the graph of f , show that ∑an converges.

Hint: See Example 4.

Consider the following grouping of terms in the harmonic series:

Show that the sum of each group of fractions is more than 1/2.Explain why this shows that the harmonic series does not converge.

Show that diverges.

Using the integral test.By considering the grouping of terms

Consider the sequence given by






(a)

(b) (c) (d)

(a)

(b)

(c)

(d)

49.

(a) (b)

(c)

50.

51.

Show that an <an +1 for all n . [Use a left-sum approximation to with Δx = 1.]

Show that an <1 for all n .Explain why limn→∞an exists.The number γ = limn→∞an is called Euler's constant. Estimate γ to two decimal places by computing a200.In the discussion following Example 4, we gave Euler's result

Find the sum of the first 20 terms of this series. Give your answer to three decimal places.

Use your answer to estimate π . Give your answer to two decimal places.

Repeat parts a and b with 100 terms.

Use a right sum approximation to bound the error in approximating π 2/6 by and by

.

This problem approximates e using

Find a lower bound for e by evaluating the first five terms of the series.

Show that 1/n ! ≤ 1/2n −1 for n ≥1.Find an upper bound for e using part b.

In this problem we investigate how fast the partial sums SN = 15+25+35+�+N 5 of the divergent series

grow as N gets larger and larger. Show that











(a)

(b)

(c)

(a) (b)

(c)

52.

53.

54.

55.

SN >N 6/6 by considering the right-hand Riemann sum for with Δx = 1.

SN <( (N +1)6−1) /6 by considering the left-hand Riemann sum for with Δx = 1.

limN→∞SN / (N 6/6) = 1. We say that SN is asymptotic to N 6/6 as N goes to infinity.In 1913, the English mathematician G. H. Hardy received a letter from the then-unknown Indian mathematical

genius Srinivasa Ramanujan, and was astounded by the results itcontained.5http://en.wikipedia.org/wiki/Srinivasa_Ramanujan, page accessed April 2, 2011. In particular, Hardywas interested in Ramanujan's results involving infinite series, such as:

In this problem you will find a formula for the general term an of this series.Write a formula in terms of n for c n where c1 = 1, c2 = −5, c3 = 9, c4 = −13, … .For products of all odd or even values up to n , we use the so-called double factorial notation n ! !. For instance,

we write 7! ! = 1×3×5×7 and 8! ! = 2×4×6×8. Bydefinition,6http://mathworld.wolfram.com/DoubleFactorial.html, page accessed April 2, 2011. we assume that(−1) ! ! = 0! ! = 1! ! = 1 and that 2! ! = 2. With this notation, write a formula in terms of n for bn where

Use your answers to parts a and b, write a formula in terms of n for an where

Strengthen Your Understanding

In Problems 53-54, explain what is wrong with the statement.

The series ∑ (1/n )2 converges because the terms approach zero as n→∞ .

The integral and the series both converge to the same value, .

In Problems 55-56, give an example of:

A series with limn→∞an = 0, but such that diverges.






56.

57.

58. 59.

60.

61.

62. 63.

64. 65.

A convergent series , whose terms are all positive, such that the series is not convergent.

Decide if the statements in Problems 57-64 are true or false. Give an explanation for your answer.

is a series of nonnegative terms.

If a series converges, then the sequence of partial sums of the series also converges.If ∑ |an +bn | converges, then ∑ |an | and ∑ |bn | converge.

The series converges.

If a series ∑an converges, then the terms, an , tend to zero as n increases.

If the terms, an , of a series tend to zero as n increases, then the series ∑an converges.If ∑an does not converge and ∑bn does not converge, then ∑anbn does not converge.

If ∑anbn converges, then ∑an and ∑bn converge.Which of the following defines a convergent sequence of partial sums?

a. Each term in the sequence is closer to the last term than any two prior consecutive terms.b. Assume that the sequence of partial sums converges to a number, L . Regardless of how small a number you give

me, say ε , I can find a value of N such that the N th term of the sequence is within ε of L .c. Assume that the sequence of partial sums converges to a number, L . I can find a value of N such that all the

terms in the sequence, past the N th term, are less than L .d. Assume that the sequence of partial sums converges to a number, L . Regardless of how small a number you give

me, say ε , I can find a value of N such that all the terms in the sequence, past the N th term, are within ε of L .

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Sequences and Series - Amazon S3 · 2015-04-02 · Sequences and Series 1/14/14 8:48 PM ... Partial Sums and Convergence of Series As in Section 9.2, we define the partial sum, Sn,