lecture 22 - section 11.2 the integral test; comparison...

Integral Test

Lecture 22Section 11.2 The Integral Test; Comparison Tests

Jiwen He

Department of Mathematics, University of Houston

[email protected]://math.uh.edu/∼jiwenhe/Math1432

∞Xk=1

f (k) converges iff

Z ∞1

f (x)dx converges

Jiwen He, University of Houston Math 1432 – Section 26626, Lecture 22 April 3, 2008 1 / 17

Integral Test Integral Test Application Comparison Tests

The Integral Test

Let ak = f (k), where f is continuous, decreasing and positive on[1,∞), then

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

Since f decreases on [1,∞)n∑

1f (x)dx <

n−1∑k=1

Since f is positive on [1,∞),∞∑

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

Therefore, either both converge or both diverge.

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The Integral Test

∞∑k=1

ak converges iff

∫ ∞

1f (x)dx converges

1f (x)dx <

n−1∑k=1

ak ≤∫ ∞

1f (x)dx ≤

∞∑k=1

The p-Sereis

(The p-Sereis)

∞∑k=1

kp= 1 +

3p+ · · · converges iff p > 1.

Proof.∞∑

kpconverges iff

∫ ∞

xpdx converges.

We know that ∫ ∞

xpdx converges iff p > 1.

It follows that∞∑

kpconverges iff p > 1.

The p-Sereis

(The p-Sereis)

∞∑k=1

kp= 1 +

Proof.∞∑

kpconverges iff

∫ ∞

xpdx converges.

The p-Sereis

(The p-Sereis)

∞∑k=1

kp= 1 +

Proof.∞∑

kpconverges iff

∫ ∞

xpdx converges.

The p-Sereis

(The p-Sereis)

∞∑k=1

kp= 1 +

Proof.∞∑

kpconverges iff

∫ ∞

xpdx converges.

The p-Sereis

(The p-Sereis)

∞∑k=1

kp= 1 +

Proof.∞∑

kpconverges iff

∫ ∞

xpdx converges.

Example (p = 1): Harmonic Series

(The p-Sereis)

∞∑k=1

kp= 1 +

Harmonic Series (p = 1)

When p = 1, this is the harmonic series

∞∑k=1

k= 1 +

3+ · · · ,

and the related improper integral is∫ ∞

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

Since this impproper integral diverges, so does the harmonic series.

(The p-Sereis)

∞∑k=1

kp= 1 +

∞∑k=1

k= 1 +

3+ · · · ,

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

(The p-Sereis)

∞∑k=1

kp= 1 +

∞∑k=1

k= 1 +

3+ · · · ,

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

(The p-Sereis)

∞∑k=1

kp= 1 +

∞∑k=1

k= 1 +

3+ · · · ,

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

(The p-Sereis)

∞∑k=1

kp= 1 +

∞∑k=1

k= 1 +

3+ · · · ,

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

(The p-Sereis)

∞∑k=1

kp= 1 +

∞∑k=1

k= 1 +

3+ · · · ,

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

(The p-Sereis)

∞∑k=1

kp= 1 +

∞∑k=1

k= 1 +

3+ · · · ,

xdx = ln x

∣∣∞1

= limx→∞

ln x = ∞.

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

When p = 2, this is∞∑

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

By the integral test,∞∑

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example (p = 2)

(The p-Sereis)

∞∑k=1

kp= 1 +

(p = 2)

k2= 1 +

32+ · · · ,

x2dx = −x−1

∣∣∞1

= 1− limx→∞

x−1 = 1.

k2= 1 +

∞∑k=2

k2≤ 1 +

∫ ∞

x2dx = 2

Example

Show that∞∑

k ln kdiverges.

The related improper integral is∫ ∞

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Since this impproper integral diverges, so does the infinite series.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

Example

Show that∞∑

k ln kdiverges.

x ln xdx =

∫ ∞

udu = ln u

∣∣∞ln 2

= limx→∞

ln x − ln ln 2 = ∞.

Show that∞∑

k(ln k)2converges.

x(ln x)2dx =

∫ ∞

u2du = −u−1

∣∣∞ln 2

ln 2− lim

x→∞x−1 =

The Integral Test for the nth Remainder∑∞

k=n+1 ak , n ≥ 1

Let ak = f (k), where f is continuous, decreasing and positive on[n,∞), then ∞∑

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

or equivalently∫ ∞

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

k=1 ak be convergent. Its nth remainder is defined as

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

Rn =∞∑

ak =∞∑

ak −n∑

ak =∞∑

ak − sn

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

k=1 ak be convergent, so do∫∞1 f (x)dx . Then, ∀ε > 0, ∃N

s.t. for n > N,∫∞n f (x)dx < ε, and

0 < Rn =∞∑

ak =∞∑

ak −n∑

ak < ε

k=n+1 ak , n ≥ 1

ak ≤∫ ∞

nf (x)dx ≤ an +

∞∑k=n+1

n+1f (x)dx ≤

∞∑k=n+1

ak ≤∫ ∞

nf (x)dx

Let∑∞

k=1 ak be convergent, so do∫∞1 f (x)dx . Then, ∀ε > 0, ∃N

s.t. for n > N,∫∞n f (x)dx < ε, and

0 < Rn =∞∑

ak =∞∑

ak −n∑

ak < ε

Example

Use a partial sum to approximate∞∑

k2 + 1with an error < 0.01

Estimating the nth remainder of the series

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

For π2 − tan−1 n < 0.01, we need n ≥ tan(π

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

Therefore1.066 . . . <

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Example

Rn =∞∑

k2 + 1−

n∑k=1

k2 + 1<

∫ ∞

x2 + 1dx

= tan−1 x∣∣∞n

2− tan−1 n.

2 − 0.01) ≈ 100.

sn =n∑

k2 + 1=

10+ · · · 1

10001= 1.066 . . .

∞∑k=1

k2 + 1< 1.076 . . .

Basic Series that Converge or Diverge

∞∑k=1

ak converges iff∞∑

ak converges, ∀j ≥ 1.

In determining whether a series converges, it does not matter wherethe summation begins. Thus, we will omit it and write

∑ak .

Basic Series that Converge

Geometric series:∑

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Basic Series that Diverge

Any series∑

ak for which limk→∞

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

∞∑k=1

∑ak .

xk , if |x | < 1

p-series:∑ 1

kp, if p > 1

Any series∑

ak 6= 0

p-series:∑ 1

kp, if p ≤ 1

Basic Comparison Test

Suppose that 0 ≤ ak ≤ bk for sufficiently large k.

bk converges, then so does∑

ak diverges, then so does∑

Comparison with p-Series∑ak converges if 0 ≤ ak ≤

kp, p > 1.∑

ak diverges if 0 ≤ c

kp≤ ak , p ≤ 1.

Comparison with Geometric Series∑ak converges if 0 ≤ ak ≤ c xk , |x | < 1.

kp, p > 1.∑

kp≤ ak , p ≤ 1.

kp, p > 1.∑

kp≤ ak , p ≤ 1.

kp, p > 1.∑

kp≤ ak , p ≤ 1.

kp, p > 1.∑

kp≤ ak , p ≤ 1.

kp, p > 1.∑

kp≤ ak , p ≤ 1.

kp, p > 1.∑

kp≤ ak , p ≤ 1.

Examples

2k3 + 1converges by comparison with

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7converges by comparison with

k5 + 4k4 + 7<

k2converges.

k3 − k2converges by comparison with

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

2k3 + 1<

k3converges.

∑ k3

k5 + 4k4 + 7<

k2converges.

k3 − k2<

k3 − k3/2=

k3, k ≥ 2, and

k3converges.

Examples

3k + 1diverges by comparison with

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)diverges by comparison with

ln(k + 6)

k + 6→ 0 as k →∞, ln(k + 6) < k + 6 for k large

ln(k + 6)>

k + 6for k large and

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Examples

3(k + 1)1

3k + 1>

3(k + 1)and

k + 1diverges.

ln(k + 6)

ln(k + 6)>

k + 6diverges.

Limit Comparison Test

Suppose that ak > 0 and bk > 0 for sufficiently large k, and that

limk→∞

bk= L for some L > 0.∑

ak converges iff∑

bk converges.

limk→∞

bk= L means that ak ≈ Lbk for large k

If limk→∞

bk= 0, and

bk converges, then∑

ak converges.

If limk→∞

bk= ∞, and

ak converges, then∑

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

limk→∞

ak converges iff∑

bk converges.

limk→∞

If limk→∞

bk= 0, and

ak converges.

If limk→∞

bk= ∞, and

bk converges.

Example

k3 − 1converges by comparison with

For large k,1

k3 − 1differs little from

k3 − 1÷ 1

k3 − 1→ 1 as k →∞

and ∑ 1

k3converges.

Example

For large k,1

k3 − 1÷ 1

k3 − 1→ 1 as k →∞

and ∑ 1

k3converges.

Example

For large k,1

k3 − 1÷ 1

k3 − 1→ 1 as k →∞

and ∑ 1

k3converges.

Example

For large k,1

k3 − 1÷ 1

k3 − 1→ 1 as k →∞

and ∑ 1

k3converges.

Example

For large k,1

k3 − 1÷ 1

k3 − 1→ 1 as k →∞

and ∑ 1

k3converges.

Example

∑ 3k2 + 2k + 1

k3 + 1diverges by comparison with

For large k,3k2 + 2k + 1

k3 + 1differs little from

3k2 + 2k + 1

k3 + 1÷ 3

3k3 + 2k2 + k

3k3 + 3→ 1 as k →∞

and ∑ 3

kdiverges.

Example

∑ 3k2 + 2k + 1

3k2 + 2k + 1

k3 + 1÷ 3

3k3 + 2k2 + k

3k3 + 3→ 1 as k →∞

and ∑ 3

kdiverges.

Example

∑ 3k2 + 2k + 1

3k2 + 2k + 1

k3 + 1÷ 3

3k3 + 2k2 + k

3k3 + 3→ 1 as k →∞

and ∑ 3

kdiverges.

Example

∑ 3k2 + 2k + 1

3k2 + 2k + 1

k3 + 1÷ 3

3k3 + 2k2 + k

3k3 + 3→ 1 as k →∞

and ∑ 3

kdiverges.

Example

∑ 3k2 + 2k + 1

3k2 + 2k + 1

k3 + 1÷ 3

3k3 + 2k2 + k

3k3 + 3→ 1 as k →∞

and ∑ 3

kdiverges.

Example

∑ 5√

k + 100

2k2√

k − 9√

kconverges by comparison with

For large k,5√

k + 100

2k2√

k − 9√

kdiffers little from

2k2√

k + 100

2k2√

k − 9√

10k2√

k + 200k2

10k2√

k − 45√

k→ 1 as k →∞

and ∑ 5

2k2converges.

Example

∑ 5√

k + 100

2k2√

k − 9√

For large k,5√

k + 100

2k2√

k − 9√

2k2√

k + 100

2k2√

k − 9√

10k2√

k + 200k2

10k2√

k − 45√

k→ 1 as k →∞

and ∑ 5

2k2converges.

Example

∑ 5√

k + 100

2k2√

k − 9√

For large k,5√

k + 100

2k2√

k − 9√

2k2√

k + 100

2k2√

k − 9√

10k2√

k + 200k2

10k2√

k − 45√

k→ 1 as k →∞

and ∑ 5

2k2converges.

Example

∑ 5√

k + 100

2k2√

k − 9√

For large k,5√

k + 100

2k2√

k − 9√

2k2√

k + 100

2k2√

k − 9√

10k2√

k + 200k2

10k2√

k − 45√

k→ 1 as k →∞

and ∑ 5

2k2converges.

Example

∑ 5√

k + 100

2k2√

k − 9√

For large k,5√

k + 100

2k2√

k − 9√

2k2√

k + 100

2k2√

k − 9√

10k2√

k + 200k2

10k2√

k − 45√

k→ 1 as k →∞

and ∑ 5

2k2converges.

Outline

The Integral TestThe Integral TestApplying the Integral TestComparison Tests

lecture 22 - section 11.2 the integral test; comparison...

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