12 INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES • In general, it is difficult to find the exact sum of a series. • We were able to accomplish this for geometric series and the series ∑ 1/[n(n+1)]. • This is because, in each of these cases, we can find a simple formula for the nth partial sum sn. • Nevertheless, usually, it isn’t easy to compute
INFINITE SEQUENCES AND SERIES • So, in the next few sections, we develop several tests that help us determine whether a series is convergent or divergent without explicitly finding its sum. • In some cases, however, our methods will enable us to find good estimates of the sum.
INFINITE SEQUENCES AND SERIES • Our first test involves improper integrals.
INFINITE SEQUENCES AND SERIES 12.3 The Integral Test and Estimates of Sums • In this section, we will learn how to: • Find the convergence or divergence of • a series and estimate its sum.
INTEGRAL TEST • We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers: • There’s no simple formula for the sum snof the first n terms.
INTEGRAL TEST • However, the computer-generated values given here suggest that the partial sums are approaching near 1.64 as n → ∞. • So, it looks as if the series is convergent. • We can confirm this impression with a geometric argument. p. 733
INTEGRAL TEST • This figure shows the curve y = 1/x2and rectangles that lie below the curve. Fig. 12.3.1, p. 733
INTEGRAL TEST • The base of each rectangle is an interval of length 1. • The height is equal to the value of the function y = 1/x2 at the right endpoint of the interval. Fig. 12.3.1, p. 733
INTEGRAL TEST • Thus, the sum of the areas of the rectangles is: Fig. 12.3.1, p. 733
INTEGRAL TEST • If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve y = 1/x2 for x≥ 1, which is the value of the integral • In Section 7.8, we discovered that this improper integral is convergent and has value 1.
INTEGRAL TEST • So, the image shows that all the partial sums are less than • Therefore, the partial sums are bounded.
INTEGRAL TEST • We also know that the partial sums are increasing (as all the terms are positive). • Thus, the partial sums converge (by the Monotonic Sequence Theorem). • So, the series is convergent.
INTEGRAL TEST • The sum of the series (the limit of the partial sums) is also less than 2:
INTEGRAL TEST • The exact sum of this series was found by the mathematician Leonhard Euler (1707–1783) to be π2/6. • However, the proof of this fact is quite difficult. • See Problem 6 in the Problems Plus, following Chapter 15.
INTEGRAL TEST • Now, let’s look at this series:
INTEGRAL TEST • The table of values of snsuggests that the partial sums aren’t approaching a finite number. • So, we suspect that the given series may be divergent. • Again, we use a picture for confirmation. p. 734
INTEGRAL TEST • The figure shows the curve y = 1/ . • However, this time, we use rectangles whose tops lie above the curve. Fig. 12.3.2, p. 734
INTEGRAL TEST • The base of each rectangle is an interval of length 1. • The height is equal to the value of the function y = 1/ at the left endpoint of the interval. Fig. 12.3.2, p. 734
INTEGRAL TEST • So, the sum of the areas of all the rectangles is: Fig. 12.3.2, p. 734
INTEGRAL TEST • This total area is greater than the area under the curve y = 1/ for x≥ 1, which is equal to the integral • However, we know from Section 7.8 that this improper integral is divergent. • In other words, the area under the curve is infinite.
INTEGRAL TEST • Thus, the sum of the series must be infinite. • That is, the series is divergent.
INTEGRAL TEST • The same sort of geometric reasoning that we used for these two series can be used to prove the following test. • The proof is given at the end of the section.
THE INTEGRAL TEST • Suppose f is a continuous, positive, decreasing function on [1, ∞) and let an =f(n). • Then, the series is convergent if and only if the improper integral is convergent.
THE INTEGRAL TEST • In other words, • If is convergent, then is convergent. • If is divergent, then is divergent.
NOTE • When we use the Integral Test, it is not necessary to start the series or the integral at n = 1. • For instance, in testing the series we use
NOTE • Also, it is not necessary that f be always decreasing. • What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N. • Then, is convergent. • So, is convergent by Note 4 of Section 11.2
INTEGRAL TEST Example 1 • Test the series for convergence or divergence. • The function f(x) = 1/(x2 + 1) is continuous, positive, and decreasing on [1, ∞).
INTEGRAL TEST Example 1 • So, we use the Integral Test:
INTEGRAL TEST Example 1 • So, is a convergent integral. • Thus, by the Integral Test, the series ∑ 1/(n2 + 1) is convergent.
INTEGRAL TEST Example 2 • For what values of p is the series convergent?
INTEGRAL TEST Example 2 • If p < 0, then • If p = 0, then • In either case, • So, the given series diverges by the Test for Divergence (Section 11.2).
INTEGRAL TEST Example 2 • If p > 0, then the function f(x) = 1/xp is clearly continuous, positive, and decreasing on [1, ∞).
INTEGRAL TEST Example 2 • In Section 7.8 (Definition 2), we found that : • Converges if p > 1 • Diverges if p ≤ 1
INTEGRAL TEST Example 2 • It follows from the Integral Test that the series ∑ 1/npconverges if p > 1 and diverges if 0 <p ≤ 1. • For p = 1, this series is the harmonic series discussed in Example 7 in Section 11.2
INTEGRAL TEST • To use the Integral Test, we need to be able to evaluate • Therefore, we have to be able to find an antiderivative of f. • Frequently, this is difficult or impossible. • So, we need other tests for convergence too.
p-SERIES • The series in Example 2 is called the p-series. • It is important in the rest of this chapter. • So, we summarize the results of Example 2 for future reference—as follows.
p-SERIES Result 1 • The p-series is convergent if p > 1 and divergent if p ≤ 1
p-SERIES Example 3 a • The series • is convergent because it is a p-series with p = 3 > 1
p-SERIES Example 3 b • The series • is divergent because it is a p-series with p = ⅓ < 1.
NOTE • We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. • In fact, whereas • Thus, in general,
INTEGRAL TEST Example 4 • Determine whether the series converges or diverges. • The function f(x) = (ln x)/x is positive and continuous for x > 1 because the logarithm function is continuous. • However, it is not obvious that f is decreasing.
INTEGRAL TEST Example 4 • So, we compute its derivative: • Thus, f’(x) < 0 when ln x > 1, that is, x > e. • It follows that f is decreasing when x > e.
INTEGRAL TEST Example 4 • So, we can apply the Integral Test: • Since this improper integral is divergent, the series Σ (ln n)/nis also divergent by the Integral Test.
ESTIMATING THE SUM OF A SERIES • Suppose we have been able to use the Integral Test to show that a series ∑ anis convergent. • Now, we want to find an approximation to the sum s of the series.s
ESTIMATING THE SUM OF A SERIES • Of course, any partial sum snis an approximation to s because • How good is such an approximation?
ESTIMATING THE SUM OF A SERIES • To find out, we need to estimate the size of the remainder Rn =s –sn =an+1 + an+2 + an+3+ … • The remainder Rnis the error made when sn, the sum of the first n terms, is used as an approximation to the total sum.
ESTIMATING THE SUM OF A SERIES • We use the same notation and ideas as in the Integral Test, assuming that fis decreasing on [n, ∞).
ESTIMATING THE SUM OF A SERIES • Comparing the areas of the rectangles with the area under y =f(x) for x >n in the figure, we see that: Fig. 12.3.3, p. 737
ESTIMATING THE SUM OF A SERIES • Similarly, from this figure, we see that: Fig. 12.3.4, p. 737