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  • Use the divergence test to determine whether a series converges or diverges.
  • Use the integral test to determine the convergence of a series.
  • Estimate the value of a series by finding bounds on its remainder term.

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums { S k } . In practice, explicitly calculating this limit can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or divergence for many types of series. In this section, we discuss two of these tests: the divergence test and the integral test. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.

Divergence test

For a series n = 1 a n to converge, the n th term a n must satisfy a n 0 as n .

Therefore, from the algebraic limit properties of sequences,

lim k a k = lim k ( S k S k 1 ) = lim k S k lim k S k 1 = S S = 0 .

Therefore, if n = 1 a n converges, the n th term a n 0 as n . An important consequence of this fact is the following statement:

If a n 0 as n , n = 1 a n diverges .

This test is known as the divergence test    because it provides a way of proving that a series diverges.

Divergence test

If lim n a n = c 0 or lim n a n does not exist, then the series n = 1 a n diverges.

It is important to note that the converse of this theorem is not true. That is, if lim n a n = 0 , we cannot make any conclusion about the convergence of n = 1 a n . For example, lim n 0 ( 1 / n ) = 0 , but the harmonic series n = 1 1 / n diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if a n 0 , the divergence test is inconclusive.

Using the divergence test

For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.

  1. n = 1 n 3 n 1
  2. n = 1 1 n 3
  3. n = 1 e 1 / n 2
  1. Since n / ( 3 n 1 ) 1 / 3 0 , by the divergence test, we can conclude that
    n = 1 n 3 n 1

    diverges.
  2. Since 1 / n 3 0 , the divergence test is inconclusive.
  3. Since e 1 / n 2 1 0 , by the divergence test, the series
    n = 1 e 1 / n 2

    diverges.
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What does the divergence test tell us about the series n = 1 cos ( 1 / n 2 ) ?

The series diverges.

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Integral test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums { S k } and showing that S 2 k > 1 + k / 2 for all positive integers k . In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test    , compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

To illustrate how the integral test works, use the harmonic series as an example. In [link] , we depict the harmonic series by sketching a sequence of rectangles with areas 1 , 1 / 2 , 1 / 3 , 1 / 4 ,… along with the function f ( x ) = 1 / x . From the graph, we see that

Practice Key Terms 4

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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