周次: 13 日期: May 20, 2022 节次: 2

In this lecture, the Comparison Test is upgraded to the limit case. In addition, two new tests based on the comparison with respect to the geometric series is given. These are called the Ratio Test and the Root Test, respectively. Note that all the tests are designed for a series with non-negative terms.

§ 11.4 The Limit Comparison Test

Note in this lecture all the series are considered to have all the terms being non-negative.

Recap: Theorem 10. The Comparison Test

  1. If there is a convergent series ncn such that

    ancn(,nN)

    then nan also converges.

  2. If there is a divergent series ndn such that

    andn(,nN)

    then nan also diverges.

Ideas behind the Proof.

For case 1, the partial sum

k=1nakk=1nckk=1ck<.

For case 2, the partial sum

k=1nakk=1ndkk=1dk=.

Example 1. Applying the Comparison Test

(a) n=155n1

( Diverges. Compared with Harmonic series n1n. Because 55n11n. )

(b) n=01n!

( Converges. Compared with the Telescoping series n=21n(n1). Because 1n!1n(n1) for all n2. Alternatively, one can compare it with the geometric series n12n1 )

(c) 5+23+17+1+n=112n+n

( Converges. Compared with the geometric series 12n+n12n)

Remarks

  • The p-series and geometric , telescoping series are frequently been used to be compared with.

  • The Comparison Test ONLY WORKS for the series with non-negative terms (or with non-positive terms).

    For example, the following series with both positive and negative terms

    n=0an=12+124+1226+1238+

    diverges, although all the terms an12n.

  • That’s been said, if a series has only finite many negative terms, the comparison test is still applicable. For example

23+n=31n2lnn

The Limit Comparison Test

Based on the experiences built up from the comparison test, we may get the feeling that testing for convergence is largely about recognizing how fast the nth term is approaching 0.

Therefore, the concepts of growth rate or order can be applied to the convergence test.

Theorem 11. Limit Comparison Test

Suppose that an>0 and bn>0 for all nN (N an integer).

  1. If

    limnanbn=c>0(an,and bn have the same order)

    then nan and nbn both converge or both diverge.

  2. If

    limnanbn=0,(an is much smaller)andnbn,converges

    then nan converges.

  3. If

    limnanbn=,(an is much bigger)andnbn,diverges

    then nan diverges.

Proof.

Example 2. Using the Limit Comparison Test

Which of the following series converges, and which diverge?

34+59+716++2n+1(n+1)2+

b.

11+13+17+115++12n1+

c.

1+2ln29+1+3ln314+1+4ln421++1+nlnnn2+5+

Example 3. Does n=1lnnn3/2 converge?

§11.5 The Ratio and Root Tests

The Ratio Test

For a geometric series narn, the ratio is a constant

an+1anr,narn converges iff |r|<1.

For other series, it appears that we can measure the rate of growth (or decline) by extending the “Ratio Test”.

Theorem 12. The Ratio Test

Let nan be a series with positive terms and suppose that

limnan+1an=ρ.

Then

(a) the series converges if ρ<1,

(b) the series diverges if ρ>1,

(c) the test is inconclusive if ρ=1.

Proof.

Example 1. Applying the Ratio Test

(a) n=02n+53n

( Converges )

(b) n=1(2n)!n!n!

( Diverges )

(c) n=14nn!n!(2n)!

( Diverges, but inconclusive from the Ratio Test )

The Root Test

The ratio test is convenient for the case when an+1/an is relatively simple. For more complicated series, we may need other test method. For example

Guiding Example

Let an={n/2n,n odd, 1/2n,n even, does nan converge?

This is not a geometric series, and the nth term approaches 0. So we do not know the series diverges. The Integral Test does not applicable as the sequence an is not decreasing. The Ratio Test produces

an+1an={12n,n odd n+12,n even

As n, the ratio has no limit.

A test that will answer the question is the Root Test.

Theorem 13. The Root Test

Let an be a series with an0 for nN, and suppose that

limnann=ρ.

Then

(a) the series converges if ρ<1

(b) the series diverges if ρ>1

(c) the test is inconclusive if ρ=1.

Proof.

Example 3. Applying the Root Test

(a) n=1n22n

( Converges )

(b) n=13n(1+1n)n2

( Diverges )

(c) n=1(11+n)n

( Converges )

Guiding Example (Revisited)

Let an={n/2n,n odd, 1/2n,n even, does nan converge?

( Converges )