9 Sequences and Series

9.4 Comparison Tests

In this section we will be comparing a given series with series that we know either converge or diverge.

Theorem 9.4.1 Direct Comparison Test

Let {an} and {bn} be positive sequences where anbn for all nN, for some N1.

  1. (a)

    If n=1bn converges, then n=1an converges.

  2. (b)

    If n=1an diverges, then n=1bn diverges.

margin: Note: A sequence {an} is a positive sequence if an>0 for all n. Because of Theorem 9.2.5, any theorem that relies on a positive sequence still holds true when an>0 for all but a finite number of values of n. Λ
  • Proof


    First consider the partial sums of each series.

    Sn=i=1naiandTn=i=1nbi

    Since both series have positive terms we know that

    SnSn+an+1=i=1nai+an+1=i=1n+1ai=Sn+1

    and

    TnTn+bn+1=i=1nbi+bn+1=i=1n+1bi=Tn+1

    Therefore, both of the sequences of partial sums,{Sn} and {Tn}, are increasing.

    For nN, We’re now going to split each series into two parts:

    S =i=1N-1ai T =i=1N-1bi
    S¯n =i=Nnai T¯n =i=Nnbi.

    This means that Sn=S+S¯n and Tn=T+T¯n. Also, because anbn for all nN, we must have S¯nT¯n.

    For the first part of the theorem, assume that n=1bn converges. Since bn0 we know that

    Tn=i=1nbii=1bi

    From above we know that S¯nT¯n for all nN so we also have

    Sn=S+S¯nS+T¯n=S+Tn-T=S-T+i=1bi

    Because i=1bi converges it must have a finite value and {Sn} is bounded above. We also showed that {Sn} is increasing so by Theorem 9.1.6 we know {Sn} converges and so n=1an converges.

    For the second part, assume that n=1an diverges. Because an0 we must have limnSn=. We also know that for all nN, S¯nT¯n. This means that

    limnTn=limn(T+T¯n)T+limnS¯n=T+limn(Sn-S)=T-S+limnSn=.

    Therefore, {Tn} is a divergent sequence and so i=1bn diverges. ∎

Example 9.4.1 Applying the Direct Comparison Test

Determine the convergence of n=113n+n2.

SolutionThis series is neither a geometric or p-series, but seems related. We predict it will converge, so we look for a series with larger terms that converges. (Note too that the Integral Test seems difficult to apply here.)

Since 3n<3n+n2, 13n>13n+n2 for all n1. The series n=113n is a convergent geometric series; by Theorem 9.4.1, n=113n+n2 converges.

Example 9.4.2 Applying the Direct Comparison Test

Determine the convergence of n=2n3n4-1.

SolutionWe know the Harmonic Series n=11n diverges, and it seems that the given series is closely related to it, hence we predict it will diverge.

We have n3n4-1>n3n4=1n for all n1.

The Harmonic Series, n=11n, diverges, so we conclude that n=1n3n4-1 diverges as well.

The concept of direct comparison is powerful and often relatively easy to apply. Practice helps one develop the necessary intuition to quickly pick a proper series with which to compare. However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test.

Consider n=1n3n4+1. It is very similar to the divergent series given in Example 9.4.2. We suspect that it also diverges, as 1nn3n4+1 for large n. However, the inequality that we naturally want to use “goes the wrong way”: since n3n4+1<n3n4=1n for all n1. The given series has terms less than the terms of a divergent series, and we cannot conclude anything from this.

Fortunately, we can apply another test to the given series to determine its convergence.

Limit Comparison Test

Theorem 9.4.2 Limit Comparison Test

Let {an} and {bn} be positive sequences.

  1. (a)

    If limnanbn=L, where L is a positive real number, then n=1an and n=1bn either both converge or both diverge.

  2. (b)

    If limnanbn=0, then if n=1bn converges, then so does n=1an.

  3. (c)

    If limnanbn=, then if n=1bn diverges, then so does n=1an.

  • Proof


    1. (a)

      We have 0<L< so we can find two positive numbers, m and M such that m<L<M. Because L=limnanbn we know that for large enough n the quotient anbn must be close to L. So there must be a positive integer N such that if n>N we also have m<anbn<M. Multiply by bn and we have mbn<an<Mbn for n>N. If n=1bn diverges, then so does n=1mbn. Also since mbn<an for sufficiently large n, by the Comparison Test n=1an also diverges.
      Similarly, if n=1bn converges, then so does n=1Mbn. Since an<Mbn for sufficiently large n, by the Comparison Test n=1an also converges.

    2. (b)

      Since limnanbn=0, there is a number N>0 such that

      |anbn-0| <1 for all n>N
      an <bn since an and bn are positive

      Now since n=1bn converges, n=1an converges by the Comparison Test.

    3. (c)

      Since limnanbn=, there is a number N>0 such that

      anbn >1 for all n>N
      an >bn for all n>N

      Now since n=1bn diverges, n=1an diverges by the Comparison Test.∎

Theorem 9.4.2 is most useful when the convergence of the series from {bn} is known and we are trying to determine the convergence of the series from {an}.

We use the Limit Comparison Test in the next example to examine the series n=1n3n4+1 which motivated this new test.

Example 9.4.3 Applying the Limit Comparison Test

Determine the convergence of n=1n3n4+1 using the Limit Comparison Test.

SolutionWe compare the terms of n=1n3n4+1 to the terms of the Harmonic Sequence n=11n:

limnn3/(n4+1)1/n =limnn4n4+1=limn11+1/n4
=1.

Since the Harmonic Series diverges, we conclude that n=1n3n4+1 diverges as well.

Example 9.4.4 Applying the Limit Comparison Test

Determine the convergence of n=113n-n2

SolutionThis series is similar to the one in Example 9.4.1, but now we are considering “3n-n2” instead of “3n+n2.” This difference makes applying the Direct Comparison Test difficult.

Instead, we use the Limit Comparison Test with the series n=113n:

limn1/(3n-n2)1/3n =limn3n3n-n2
=by LHRlimnln33nln33n-2n
=by LHRlimn(ln3)23n(ln3)23n-2
=by LHRlimn(ln3)33n(ln3)33n=1.

We know n=113n is a convergent geometric series, hence n=113n-n2 converges as well.

As mentioned before, practice helps one develop the intuition to quickly choose a series with which to compare. A general rule of thumb is to pick a series based on the dominant term in the expression of {an}. It is also helpful to note that factorials dominate increasing exponentials, which dominate algebraic functions (e.g., polynomials), which dominate logarithms. In the previous example, the dominant term of 13n-n2 was 3n, so we compared the series to n=113n. It is hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L’Hôpital’s Rule to n!.

Example 9.4.5 Applying the Limit Comparison Test

Determine the convergence of n=1n+3n2-n+1.

SolutionWe naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is 1/n2. Knowing that n=11n2 converges, we attempt to apply the Limit Comparison Test:

limn(n+3)/(n2-n+1)1/n2 =limnn2(n+3)n2-n+1
=(Apply L’Hôpital’s Rule).

Theorem 9.4.2 part (3) only applies when n=1bn diverges; in our case, it converges. Ultimately, our test has not revealed anything about the convergence of our series.

The problem is that we chose a poor series with which to compare. Since the numerator and denominator of the terms of the series are both algebraic functions, we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator.

The dominant term of the numerator is n1/2 and the dominant term of the denominator is n2. Thus we should compare the terms of the given series to n1/2/n2=1/n3/2:

limn(n+3)/(n2-n+1)1/n3/2 =limnn3/2(n+3)n2-n+1
=1(Apply L’Hôpital’s Rule).

Since the p-series n=11n3/2 converges, we conclude that n=1n+3n2-n+1 converges as well.

The tests we have encountered so far have required that we analyze series from positive sequences (the absolute value of the ratio and the root tests of Section 9.6 will convert the sequence into a positive sequence). The next section relaxes this restriction by considering alternating series, where the underlying sequence has terms that alternate between being positive and negative.

Exercises 9.4

 

Terms and Concepts

  1. 1.

    Suppose n=0an is convergent, and there are sequences {bn} and {cn} such that 0bnancn for all n. What can be said about the series n=0bn and n=0cn?

  2. 2.

    Suppose n=0an is divergent, and there are sequences {bn} and {cn} such that 0bnancn for all n. What can be said about the series n=0bn and n=0cn?

Problems

In Exercises 3–8., use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison.

  1. 3.

    n=11n2+3n-5

  2. 4.

    n=114n+n2-n

  3. 5.

    n=1lnnn

  4. 6.

    n=11n!+n

  5. 7.

    n=21n2-1

  6. 8.

    n=51n-2

In Exercises 9–14., use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison.

  1. 9.

    n=11n2-3n+5

  2. 10.

    n=114n-n2

  3. 11.

    n=4lnnn-3

  4. 12.

    n=11n2+n

  5. 13.

    n=11n+n

  6. 14.

    n=1sin(1/n)

In Exercises 15–24., use the Direct Comparison Test or the Limit Comparison Test to determine the convergence of the given series. State which series is used for comparison.

  1. 15.

    n=1n2+n+1n3-5

  2. 16.

    n=1n-10n2+10n+10

  3. 17.

    n=12n5n+10

  4. 18.

    n=1n+5n3-5

  5. 19.

    n=1nn4+1

  6. 20.

    n=1n-1n4n

  7. 21.

    n=2nn2-1

  8. 22.

    n=11n+100

  9. 23.

    n=21n2lnn

  10. 24.

    n=1n+3n2+17

In Exercises 25–32., determine the convergence of the given series. State the test used; more than one test may be appropriate.

  1. 25.

    n=1n22n

  2. 26.

    n=11(2n+5)3

  3. 27.

    n=1n!10n

  4. 28.

    n=1lnnn!

  5. 29.

    n=113n+n

  6. 30.

    n=1n-210n+5

  7. 31.

    n=13nn3

  8. 32.

    n=1cos(1/n)n

  1. 33.
    Given that n=1an converges, state which of the following series converges, may converge, or does not converge. (a) n=1ann (b) n=1anan+1 (c) n=1(an)2 (d) n=1nan (e) n=11an
  2. 34.

    We have shown that the harmonic series 1+12+13+14+ diverges. Suppose we remove some terms by considering the series n=11pn where pn is the nth prime (so p1=2, p2=3, p3=5, etc). Determine if this series converges or diverges, using the fact that pn<2nlnn for all n sufficiently large.

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