INFINITE SEQUENCES AND SERIES

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INFINITE SEQUENCES AND SERIES
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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(n1).
• This is because, in each of these cases, we can
  find a simple formula for the nth partial sum
  sn.
• Nevertheless, usually, it isnt easy to compute

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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.

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INFINITE SEQUENCES AND SERIES

• Our first test involves improper integrals.

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INFINITE SEQUENCES AND SERIES
11.3 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.
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INTEGRAL TEST

• We begin by investigating the series whose terms
  are the reciprocals of the squares of the
  positive integers
• Theres no simple formula for the sum sn of the
  first n terms.

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INTEGRAL TEST

• However, the computer-generated values given here
  suggest that the partial sums are approaching
  near 1.64 as n ? 8.
• So, it looks as if the series is convergent.
• We can confirm this impression with a geometric
  argument.

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INTEGRAL TEST

• This figure shows the curve y 1/x2 and
  rectangles that lie below the curve.

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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.

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INTEGRAL TEST

• Thus, the sum of the areas of the rectangles is

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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.

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INTEGRAL TEST

• So, the image shows that all the partial sums are
  less than
• Therefore, the partial sums are bounded.

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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.

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INTEGRAL TEST

• The sum of the series (the limit of the partial
  sums) is also less than 2

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INTEGRAL TEST

• The exact sum of this series was found by the
  mathematician Leonhard Euler (17071783) to be
  p2/6.
• However, the proof of this fact is quite
  difficult.
• See Problem 6 in the Problems Plus, following
  Chapter 15.

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INTEGRAL TEST

• Now, lets look at this series

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INTEGRAL TEST

• The table of values of sn suggests that the
  partial sums arent approaching a finite number.
• So, we suspect that the given series may be
  divergent.
• Again, we use a picture for confirmation.

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INTEGRAL TEST

• The figure shows the curve y 1/ .
• However, this time, we use rectangles whose tops
  lie above the curve.

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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.

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INTEGRAL TEST

• So, the sum of the areas of all the rectangles is

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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.

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INTEGRAL TEST

• Thus, the sum of the series must be infinite.
• That is, the series is divergent.

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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.

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THE INTEGRAL TEST

• Suppose f is a continuous, positive, decreasing
  function on 1, 8) and let an f(n).
• Then, the series is convergent if and only
  if the improper integral is
  convergent.

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THE INTEGRAL TEST

• In other words,
• If is convergent, then is
  convergent.
• If is divergent, then is divergent.

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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

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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

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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, 8).

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INTEGRAL TEST
Example 1

• So, we use the Integral Test

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INTEGRAL TEST
Example 1

• So, is a convergent integral.
• Thus, by the Integral Test, the series ? 1/(n2
  1) is convergent.

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INTEGRAL TEST
Example 2

• For what values of p is the series
  convergent?

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INTEGRAL TEST
Example 2

• If p lt 0, then
• If p 0, then
• In either case,
• So, the given series diverges by the Test for
  Divergence (Section 11.2).

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INTEGRAL TEST
Example 2

• If p gt 0, then the function f(x) 1/xp is
  clearly continuous, positive, and decreasing on
  1, 8).

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INTEGRAL TEST
Example 2

• In Section 7.8 (Definition 2), we found that
  
• Converges if p gt 1
• Diverges if p 1

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INTEGRAL TEST
Example 2

• It follows from the Integral Test that the
  series ? 1/np converges if p gt 1 and diverges if
  0 lt p 1.
• For p 1, this series is the harmonic series
  discussed in Example 7 in Section 11.2

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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.

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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 referenceas follows.

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p-SERIES
Result 1

• The p-series is convergent if p gt 1 and
  divergent if p 1

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p-SERIES
Example 3 a

• The series
• is convergent because it is a p-series with p
  3 gt 1

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p-SERIES
Example 3 b

• The series
• is divergent because it is a p-series with p ?
  lt 1.

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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,

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INTEGRAL TEST
Example 4

• Determine whether the series converges or
  diverges.
• The function f(x) (ln x)/x is positive and
  continuous for x gt 1 because the logarithm
  function is continuous.
• However, it is not obvious that f is decreasing.

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INTEGRAL TEST
Example 4

• So, we compute its derivative
• Thus, f(x) lt 0 when ln x gt 1, that is, x gt e.
• It follows that f is decreasing when x gt e.

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INTEGRAL TEST
Example 4

• So, we can apply the Integral Test
• Since this improper integral is divergent, the
  series S (ln n)/n is also divergent by the
  Integral Test.

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ESTIMATING THE SUM OF A SERIES

• Suppose we have been able to use the Integral
  Test to show that a series ? an is convergent.
• Now, we want to find an approximation to the sum
  s of the series.s

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ESTIMATING THE SUM OF A SERIES

• Of course, any partial sum sn is an
  approximation to s because
• How good is such an approximation?

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ESTIMATING THE SUM OF A SERIES

• To find out, we need to estimate the size of the
  remainder Rn s sn an1 an2 an3
• The remainder Rn is the error made when sn, the
  sum of the first n terms, is used as an
  approximation to the total sum.

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ESTIMATING THE SUM OF A SERIES

• We use the same notation and ideas as in the
  Integral Test, assuming that f is decreasing on
  n, 8).

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ESTIMATING THE SUM OF A SERIES

• Comparing the areas of the rectangles with the
  area under y f(x) for x gt n in the figure, we
  see that

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ESTIMATING THE SUM OF A SERIES

• Similarly, from this figure, we see that

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ESTIMATING THE SUM OF A SERIES

• Thus, we have proved the following error
  estimate.

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REMAINDER ESTIMATE (INT. TEST)
Estimate 2

• Suppose f(k) ak, where f is a continuous,
  positive, decreasing function for x n and ? an
  is convergent.
• If Rn s sn, then

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REMAINDER ESTIMATE
Example 5

1. Approximate the sum of the series ? 1/n3 by
   using the sum of the first 10 terms. Estimate
   the error involved.
2. How many terms are required to ensure the sum is
   accurate to within 0.0005?

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REMAINDER ESTIMATE
Example 5

• In both parts, we need to know
• With f(x) 1/x3, which satisfies the conditions
  of the Integral Test, we have

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REMAINDER ESTIMATE
Example 5 a

• As per the remainder estimate 2, we have
• So, the size of the error is at most 0.005

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REMAINDER ESTIMATE
Example 5 b

• Accuracy to within 0.0005 means that we have to
  find a value of n such that Rn 0.0005
• Since
• we want

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REMAINDER ESTIMATE
Example 5 b

• Solving this inequality, we get
• We need 32 terms to ensure accuracy to within
  0.0005

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REMAINDER ESTIMATE
Estimate 3

• If we add sn to each side of the inequalitiesin
  Estimate 2, we get
• because sn Rn s

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REMAINDER ESTIMATE

• The inequalities in Estimate 3 give a lower bound
  and an upper bound for s.
• They provide a more accurate approximation to
  the sum of the series than the partial sum sn
  does.

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REMAINDER ESTIMATE
Example 6

• Use Estimate 3 with n 10 to estimate the sum of
  the series

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REMAINDER ESTIMATE
Example 6

• The inequalities in Estimate 3 become

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REMAINDER ESTIMATE
Example 6

• From Example 5, we know that
• Thus,

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REMAINDER ESTIMATE
Example 6

• Using s10 1.197532, we get
• 1.201664 s 1.202532
• If we approximate s by the midpoint of this
  interval, then the error is at most half the
  length of the interval.
• Thus,

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REMAINDER ESTIMATE

• If we compare Example 6 with Example 5, we see
  that the improved estimate 3 can be much better
  than the estimate s sn.
• To make the error smaller than 0.0005, we had to
  use 32 terms in Example 5, but only 10 terms in
  Example 6.

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PROOF OF THE INTEGRAL TEST

• We have already seen the basic idea behind the
  proof of the Integral Test for the series ? 1/n2
  and ? 1/ .

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PROOF OF THE INTEGRAL TEST

• For the general series ? an, consider these
  figures.

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PROOF OF THE INTEGRAL TEST

• The area of the first shaded rectangle in this
  figure is the value of f at the right endpoint of
  1, 2, that is, f(2) a2.

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PROOF OF THE INTEGRAL TEST
Estimate 4

• So, comparing the areas of the rectangles with
  the area under y f(x) from 1 to n, we see
  that
• Notice that this inequality depends on the fact
  that f is decreasing.

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PROOF OF THE INTEGRAL TEST
Estimate 5

• Likewise, the figure shows that

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PROOF OF THE INTEGRAL TEST
Case i

• If is convergent, then Estimate 4
  gives
• since f(x) 0.

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PROOF OF THE INTEGRAL TEST
Case i

• Therefore,
• Since sn M for all n, the sequence sn is
  bounded above.

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PROOF OF THE INTEGRAL TEST
Case i

• Also,
• since an1 f(n 1) 0.
• Thus, sn is an increasing bounded sequence.

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PROOF OF THE INTEGRAL TEST
Case i

• Thus, it is convergent by the Monotonic Sequence
  Theorem (Section 11.1).
• This means that ? an is convergent.

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PROOF OF THE INTEGRAL TEST
Case ii

• If is divergent, then as
  n ? 8 because f(x) 0.

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PROOF OF THE INTEGRAL TEST
Case ii

• However, Estimate 5 gives
• Hence, sn1?8.
• This implies that sn ? 8, and so ? an diverges.