lecture 26 : comparison testapilking/math10560/lectures/lecture... · 2020-01-13 · comparison...

Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example Example

Comparison Test

In this section, as we did with improper integrals, we see how to compare aseries (with Positive terms) to a well known series to determine if it convergesor diverges.

I We will of course make use of our knowledge of p-series and geometricseries.

∞Xn=1

npconverges for p > 1, diverges for p ≤ 1.

∞Xn=1

arn−1 converges if |r | < 1, diverges if |r | ≥ 1.

I Comparison Test Suppose thatP

an andP

bn are series with positiveterms.

(i) IfP

bn is convergent and an ≤ bn for all n, thanP

an is alsoconvergent.

(ii) IfP

bn is divergent and an ≥ bn for all n, thenP

an is divergent.

Annette Pilkington Lecture 26 : Comparison Test

Comparison Test

∞Xn=1

an andP

(i) IfP

(ii) IfP

an is divergent.

Comparison Test

∞Xn=1

an andP

(i) IfP

(ii) IfP

an is divergent.

Example 1

Example 1 Use the comparison test to determine if the following seriesconverges or diverges:

∞Xn=1

2−1/n

I First we check that an > 0 –> true since 2−1/n

n3 > 0 for n ≥ 1.

I We have 21/n = n√

2 > 1 for n ≥ 1. Therefore 2−1/n = 1n√2< 1 for n ≥ 1.

I Therefore 2−1/n

n3 < 1n3 for n > 1.

I SinceP∞

n=11n3 is a p-series with p > 1, it converges.

I Comparing the above series withP∞

n=11n3 , we can conclude thatP∞

n=12−1/n

n3 also converges andP∞

n=12−1/n

n3 ≤P∞

n=11n3

Example 1

∞Xn=1

2−1/n

n3 > 0 for n ≥ 1.

I Therefore 2−1/n

n3 < 1n3 for n > 1.

I SinceP∞

n=12−1/n

n3 ≤P∞

n=11n3

Example 1

∞Xn=1

2−1/n

n3 > 0 for n ≥ 1.

I Therefore 2−1/n

n3 < 1n3 for n > 1.

I SinceP∞

n=12−1/n

n3 ≤P∞

n=11n3

Example 1

∞Xn=1

2−1/n

n3 > 0 for n ≥ 1.

I Therefore 2−1/n

n3 < 1n3 for n > 1.

I SinceP∞

n=12−1/n

n3 ≤P∞

n=11n3

Example 1

∞Xn=1

2−1/n

n3 > 0 for n ≥ 1.

I Therefore 2−1/n

n3 < 1n3 for n > 1.

I SinceP∞

n=12−1/n

n3 ≤P∞

n=11n3

Example 1

∞Xn=1

2−1/n

n3 > 0 for n ≥ 1.

I Therefore 2−1/n

n3 < 1n3 for n > 1.

I SinceP∞

n=12−1/n

n3 ≤P∞

n=11n3

Example 2

∞Xn=1

I First we check that an > 0 –> true since 21/n

n> 0 for n ≥ 1.

2 > 1 for n ≥ 1.

I Therefore 21/n

nfor n > 1.

I SinceP∞

is a p-series with p = 1 (a.k.a. the harmonic series), itdiverges.

I Therefore, by comparison, we can conclude thatP∞

n=121/n

nalso diverges.

Example 2

∞Xn=1

n> 0 for n ≥ 1.

2 > 1 for n ≥ 1.

I Therefore 21/n

nfor n > 1.

I SinceP∞

n=121/n

nalso diverges.

Example 2

∞Xn=1

n> 0 for n ≥ 1.

2 > 1 for n ≥ 1.

I Therefore 21/n

nfor n > 1.

I SinceP∞

n=121/n

nalso diverges.

Example 2

∞Xn=1

n> 0 for n ≥ 1.

2 > 1 for n ≥ 1.

I Therefore 21/n

nfor n > 1.

I SinceP∞

n=121/n

nalso diverges.

Example 2

∞Xn=1

n> 0 for n ≥ 1.

2 > 1 for n ≥ 1.

I Therefore 21/n

nfor n > 1.

I SinceP∞

n=121/n

nalso diverges.

Example 2

∞Xn=1

n> 0 for n ≥ 1.

2 > 1 for n ≥ 1.

I Therefore 21/n

nfor n > 1.

I SinceP∞

n=121/n

nalso diverges.

Example 3

∞Xn=1

n2 + 1

I First we check that an > 0 –> true since 1n2+1

> 0 for n ≥ 1.

I We have n2 + 1 > n2 for n ≥ 1.

I Therefore 1n2+1

< 1n2 for n > 1.

I SinceP∞

n=11n2 is a p-series with p = 2, it converges.

n2+1also converges

andP∞

n2+1≤P∞

n=11n2 .

Example 3

∞Xn=1

n2 + 1

> 0 for n ≥ 1.

I Therefore 1n2+1

< 1n2 for n > 1.

I SinceP∞

n2+1also converges

andP∞

n2+1≤P∞

n=11n2 .

Example 3

∞Xn=1

n2 + 1

> 0 for n ≥ 1.

I Therefore 1n2+1

< 1n2 for n > 1.

I SinceP∞

n2+1also converges

andP∞

n2+1≤P∞

n=11n2 .

Example 3

∞Xn=1

n2 + 1

> 0 for n ≥ 1.

I Therefore 1n2+1

< 1n2 for n > 1.

I SinceP∞

n2+1also converges

andP∞

n2+1≤P∞

n=11n2 .

Example 3

∞Xn=1

n2 + 1

> 0 for n ≥ 1.

I Therefore 1n2+1

< 1n2 for n > 1.

I SinceP∞

n2+1also converges

andP∞

n2+1≤P∞

n=11n2 .

Example 3

∞Xn=1

n2 + 1

> 0 for n ≥ 1.

I Therefore 1n2+1

< 1n2 for n > 1.

I SinceP∞

n2+1also converges

andP∞

n2+1≤P∞

n=11n2 .

Example 4

∞Xn=1

I First we check that an > 0 –> true since n−2

2n = 1n22n > 0 for n ≥ 1.

I We have 1n22n <

1n2 for n ≥ 1.

I SinceP∞

n22n also convergesand

P∞n=1

1n22n ≤

P∞n=1

Example 4

∞Xn=1

2n = 1n22n > 0 for n ≥ 1.

I We have 1n22n <

1n2 for n ≥ 1.

I SinceP∞

P∞n=1

1n22n ≤

P∞n=1

Example 4

∞Xn=1

2n = 1n22n > 0 for n ≥ 1.

I We have 1n22n <

1n2 for n ≥ 1.

I SinceP∞

P∞n=1

1n22n ≤

P∞n=1

Example 4

∞Xn=1

2n = 1n22n > 0 for n ≥ 1.

I We have 1n22n <

1n2 for n ≥ 1.

I SinceP∞

P∞n=1

1n22n ≤

P∞n=1

Example 4

∞Xn=1

2n = 1n22n > 0 for n ≥ 1.

I We have 1n22n <

1n2 for n ≥ 1.

I SinceP∞

P∞n=1

1n22n ≤

P∞n=1

Example 4

∞Xn=1

2n = 1n22n > 0 for n ≥ 1.

I We have 1n22n <

1n2 for n ≥ 1.

I SinceP∞

P∞n=1

1n22n ≤

P∞n=1

Example 5

∞Xn=1

I First we check that an > 0 –> true since ln nn> 1

n> 0 for n ≥ e. Note

that this allows us to use the test since a finite number of terms have nobearing on convergence or divergence.

I We have ln nn> 1

nfor n > 3.

I SinceP∞

diverges, we can conclude thatP∞

n=1ln nn

also diverges.

Example 5

∞Xn=1

I We have ln nn> 1

nfor n > 3.

I SinceP∞

n=1ln nn

also diverges.

Example 5

∞Xn=1

I We have ln nn> 1

nfor n > 3.

I SinceP∞

n=1ln nn

also diverges.

Example 5

∞Xn=1

I We have ln nn> 1

nfor n > 3.

I SinceP∞

n=1ln nn

also diverges.

Example 6

∞Xn=1

I First we check that an > 0 –> true since 1n!> 0 for n ≥ 1.

I We have n! = n(n − 1)(n − 2) · · · · 2 · 1 > 2 · 2 · 2 · · · · · 2 · 1 = 2n−1.Therefore 1

2n−1 .

I SinceP∞

2n−1 converges, we can conclude thatP∞

n=11n!

also converges.

Example 6

∞Xn=1

2n−1 .

I SinceP∞

n=11n!

also converges.

Example 6

∞Xn=1

2n−1 .

I SinceP∞

n=11n!

also converges.

Example 6

∞Xn=1

2n−1 .

I SinceP∞

n=11n!

also converges.

Limit Comparison Test

Limit Comparison Test Suppose thatP

an andP

bn are series withpositive terms. If

limn→∞

where c is a finite number and c > 0, then either both series converge or bothdiverge. (Note c 6= 0 or ∞. )

Example Test the following series for convergence using the LimitComparison test:

∞Xn=2

n2 − 1

(Note that our previous comparison test is difficult to apply in this and most ofthe examples below.)

I First we check that an > 0 –> true since an = 1n2−1

> 0 for n ≥ 2. (afterwe study absolute convergence, we see how to get around this restriction.)

I We will compare this series toP∞

n=21n2 which converges, since it is a

p-series with p = 2. bn = 1n2 .

I limn→∞anbn

= limn→∞1/(n2−1)

1/n2 = limn→∞n2

n2−1= limn→∞

11−(1/n2)

I Since c = 1 > 0, we can conclude that both series converge.

an andP

limn→∞

∞Xn=2

n2 − 1

I limn→∞anbn

= limn→∞1/(n2−1)

1/n2 = limn→∞n2

n2−1= limn→∞

11−(1/n2)

an andP

limn→∞

∞Xn=2

n2 − 1

I limn→∞anbn

= limn→∞1/(n2−1)

1/n2 = limn→∞n2

n2−1= limn→∞

11−(1/n2)

an andP

limn→∞

∞Xn=2

n2 − 1

I limn→∞anbn

= limn→∞1/(n2−1)

1/n2 = limn→∞n2

n2−1= limn→∞

11−(1/n2)

an andP

limn→∞

∞Xn=2

n2 − 1

I limn→∞anbn

= limn→∞1/(n2−1)

1/n2 = limn→∞n2

n2−1= limn→∞

11−(1/n2)

Example

an andP

bn are series withpositive terms. If limn→∞

= c where c is a finite number and c > 0, theneither both series converge or both diverge. (Note c 6= 0 or ∞. )

∞Xn=1

n2 + 2n + 1

n4 + n2 + 2n + 1

I First we check that an > 0 –> true since an = n2+2n+1n4+n2+2n+1

> 0 for n ≥ 1.

I For a rational function, the rule of thumb is to compare the series to theseries

nq , where p is the degree of the numerator and q is the degree ofthe denominator.

n4 =P∞

n=11n2 which converges, since

it is a p-series with p = 2. bn = 1n2 .

I limn→∞anbn

= limn→∞( n2+2n+1n4+n2+2n+1

)/(1/n2) = limn→∞n4+2n3+n2

n4+n2+2n+1=

limn→∞1+2/n+1/n2

1+1/n2+2/n3+1/n4 = 1.

Example

an andP

∞Xn=1

n2 + 2n + 1

n4 + n2 + 2n + 1

> 0 for n ≥ 1.

n4 =P∞

I limn→∞anbn

= limn→∞( n2+2n+1n4+n2+2n+1

)/(1/n2) = limn→∞n4+2n3+n2

n4+n2+2n+1=

limn→∞1+2/n+1/n2

1+1/n2+2/n3+1/n4 = 1.

Example

an andP

∞Xn=1

n2 + 2n + 1

n4 + n2 + 2n + 1

> 0 for n ≥ 1.

n4 =P∞

I limn→∞anbn

= limn→∞( n2+2n+1n4+n2+2n+1

)/(1/n2) = limn→∞n4+2n3+n2

n4+n2+2n+1=

limn→∞1+2/n+1/n2

1+1/n2+2/n3+1/n4 = 1.

Example

an andP

∞Xn=1

n2 + 2n + 1

n4 + n2 + 2n + 1

> 0 for n ≥ 1.

n4 =P∞

I limn→∞anbn

= limn→∞( n2+2n+1n4+n2+2n+1

)/(1/n2) = limn→∞n4+2n3+n2

n4+n2+2n+1=

limn→∞1+2/n+1/n2

1+1/n2+2/n3+1/n4 = 1.

Example

an andP

∞Xn=1

n2 + 2n + 1

n4 + n2 + 2n + 1

> 0 for n ≥ 1.

n4 =P∞

I limn→∞anbn

= limn→∞( n2+2n+1n4+n2+2n+1

)/(1/n2) = limn→∞n4+2n3+n2

n4+n2+2n+1=

limn→∞1+2/n+1/n2

1+1/n2+2/n3+1/n4 = 1.

Example

an andP

∞Xn=1

n2 + 2n + 1

n4 + n2 + 2n + 1

> 0 for n ≥ 1.

n4 =P∞

I limn→∞anbn

= limn→∞( n2+2n+1n4+n2+2n+1

)/(1/n2) = limn→∞n4+2n3+n2

n4+n2+2n+1=

limn→∞1+2/n+1/n2

1+1/n2+2/n3+1/n4 = 1.

Example

∞Xn=1

2n + 1√n3 + 1

I First we check that an > 0 –> true since an = 2n+1√n3+1

> 0 for n ≥ 1.

n=1n√n3

n3/2 =P∞

n=11√n

diverges, since it is a p-series with p = 1/2. bn =1√n

I limn→∞anbn

= limn→∞( 2n+1√n3+1

(1/√

n) = limn→∞+2n3/2+

√n√

limn→∞(2n3/2+

√n)‹

√n3+1

‹n3/2

= limn→∞(2+1/n)√(n3+1)/n3

= limn→∞(2+1/n)√(1+1/n3)

I Since c = 2 > 0, we can conclude that both series diverge.

Example

∞Xn=1

2n + 1√n3 + 1

> 0 for n ≥ 1.

n=1n√n3

n3/2 =P∞

n=11√n

I limn→∞anbn

= limn→∞( 2n+1√n3+1

(1/√

n) = limn→∞+2n3/2+

√n√

limn→∞(2n3/2+

√n)‹

√n3+1

‹n3/2

= limn→∞(2+1/n)√(n3+1)/n3

= limn→∞(2+1/n)√(1+1/n3)

Example

∞Xn=1

2n + 1√n3 + 1

> 0 for n ≥ 1.

n=1n√n3

n3/2 =P∞

n=11√n

I limn→∞anbn

= limn→∞( 2n+1√n3+1

(1/√

n) = limn→∞+2n3/2+

√n√

limn→∞(2n3/2+

√n)‹

√n3+1

‹n3/2

= limn→∞(2+1/n)√(n3+1)/n3

= limn→∞(2+1/n)√(1+1/n3)

Example

∞Xn=1

2n + 1√n3 + 1

> 0 for n ≥ 1.

n=1n√n3

n3/2 =P∞

n=11√n

I limn→∞anbn

= limn→∞( 2n+1√n3+1

(1/√

n) = limn→∞+2n3/2+

√n√

limn→∞(2n3/2+

√n)‹

√n3+1

‹n3/2

= limn→∞(2+1/n)√(n3+1)/n3

= limn→∞(2+1/n)√(1+1/n3)

Example

∞Xn=1

2n + 1√n3 + 1

> 0 for n ≥ 1.

n=1n√n3

n3/2 =P∞

n=11√n

I limn→∞anbn

= limn→∞( 2n+1√n3+1

(1/√

n) = limn→∞+2n3/2+

√n√

limn→∞(2n3/2+

√n)‹

√n3+1

‹n3/2

= limn→∞(2+1/n)√(n3+1)/n3

= limn→∞(2+1/n)√(1+1/n3)

Example

∞Xn=1

2n + 1√n3 + 1

> 0 for n ≥ 1.

n=1n√n3

n3/2 =P∞

n=11√n

I limn→∞anbn

= limn→∞( 2n+1√n3+1

(1/√

n) = limn→∞+2n3/2+

√n√

limn→∞(2n3/2+

√n)‹

√n3+1

‹n3/2

= limn→∞(2+1/n)√(n3+1)/n3

= limn→∞(2+1/n)√(1+1/n3)

Example

∞Xn=1

2n − 1

I First we check that an > 0 –> true since an = e2n−1

> 0 for n ≥ 1.

n=112n which converges, since it is a

geometric series with r = 1/2 < 1. bn =1

I limn→∞anbn

= limn→∞( e2n−1

(1/2n) = limn→∞e

1−1/2n = e.

I Since c = e > 0, we can conclude that both series converge.

Example

∞Xn=1

2n − 1

> 0 for n ≥ 1.

I limn→∞anbn

(1/2n) = limn→∞e

1−1/2n = e.

Example

∞Xn=1

2n − 1

> 0 for n ≥ 1.

I limn→∞anbn

(1/2n) = limn→∞e

1−1/2n = e.

Example

∞Xn=1

2n − 1

> 0 for n ≥ 1.

I limn→∞anbn

(1/2n) = limn→∞e

1−1/2n = e.

Example

∞Xn=1

2n − 1

> 0 for n ≥ 1.

I limn→∞anbn

(1/2n) = limn→∞e

1−1/2n = e.

Example

∞Xn=1

I First we check that an > 0 –> true since an = 21/n

n2 > 0 for n ≥ 1.

p-series with p = 2 > 1. bn =1

I limn→∞anbn

= limn→∞( 21/n