Ch 9.5Calculus Graphical, Numerical, Algebraic byFinney, Demana, Waits, Kennedy

Testing Convergence at Endpoints

Convergence of Two Series

2n=1 n=1

1 1Consider the two series, and

n n

1. What does the ratio test show about convergence of both series?

2. Use improper integrals to show the area of both curves over the interval 1 ≤ x ≤ ∞.

3. How does this relate to the ratio test?

Convergence of Two Series

2n=1 n=1

1 1Consider the two series, and

n n

2 2

1 2 2n n n n

2

11n + 1n nn + 1

L = lim = lim = 1 L = lim = lim = 11 1n + 1 n + 1n n

3. How does this relate to the ratio test? Ratio Test is inconclusive when L = 1; but Integral Test works.

1. What does the ratio test show about convergence of both series?

2. Use improper integrals to show the area of both curves over the interval 1 ≤ x ≤ ∞.

k1k k

1

-1 k12 k k

1

1 dx = lim ln x| = lim ln k =

x

1 1 dx = lim -x | = lim - + 1 = 1

x k

Using the Ratio Test gives a limit L =1 which is inconclusive.

The p-Series Test

pn=1

pn=1

pn=1

11. Use the Integral Test to prove that converges if p > 1.

n

12. Use the Integral Test to prove that diverges if p < 1.

n

13. Use the Integral Test to prove that diverges if p

n

= 1.

The p-Series Test

pn=1

k kk -p+1

p p p 1k k k1 1 11

p-1k

11. Use the Integral Test to prove that converges if .

n

1 1 x 1 1 dx = lim dx = lim = lim

x x -p + 1 1 p x

1 1 lim - 1

1- p

p

k

> 1

pn=1

kk -p+1

p pk k1 1 1

1 1 = - =

1-p p 1

12. Use the Integral Test to prove tha

The series converges by test.

0 < p < 1t diverges if .n

1 1 x dx = lim dx = lim

x x -p + 1

1- p

k

pn=1

k

p k1 1

1= lim k - 1 =

1- p

since (1 - p > 0).

13. Use the Integral Test to prove that diverges if .

n

The series diverges by test.

p = 1

1 1 dx = lim dx =

x x

k

1k k lim ln x = lim ln k =

The series diverges by test.

Slow Divergence of Harmonic Series

n 1

n

1

Approximately

1 1 1 1 1

n 2 3 4 n

1

x

how many terms of the harmonic series

are required to form a partial sum > 20?

= 1 + + + + ... + > 20

< 1 + dx = 1 + ln n

19

1 + ln n > 20

ln n > 19

n > e 178,482,301 terms!!

Example

21 x

Determine whether the improper integral converges or diverges:

1 + cosx dx

Example

2 2 21 1 1

21

21

x x x

and

x

x

Determine whether the improper integral converges or diverges:

1 + cosx 1 + cosx 2 dx dx dx

2 dx converges since this is a p-series with p >1

Therefore,

1 + cosx d

x converges by comparison test and p-series test.

Limit Comparison Test

nn=1

Deter min e whether the series converges or diverges:

1 1 1 1 1 + + + + ... +

1 3 7 15 2 - 1

Limit Comparison Test

n n

nnn

n

n

nn=1

n

Deter min e whether the series converges or diverges:

1 1 1 1 1 + + + + ... +

1 3 7 15 2 - 1

For n large, 1/ (2 - 1) behaves like so we compare the se1/2 1/2

a 1lim

ries to

= l im b 2 -

n n

n n

nn 1

n n

n

=

Since 1/2 converges, the LCT guarantees that

1

2 2 1 = lim = lim = 1

1 1 2

also converges.2

- 1 1 - (

- 1

1/

2 )

Limit Comparison Test

n=1

Deter min e whether the series converges or diverges:

1 1 1sin 1 + sin + sin + ... + sin

2 3 n

Limit Comparison Test

n

n n

n=1

x 0

Deter min e whether the series converges or diverges:

1 1 1 sin 1 + sin + sin + ... + sin

2 3 n

sin xUsing lim = 1, it is useful to compare the series to (1/n)

x

toa

li get: mb

n

n=1

n 0

1Since (1/n) diverges, by

sin 1/n si

,

n n = lim = lim =

sin also diverges.n

11/n n

LCT

Alternating Harmonic Series

Prove that the alternating harmonic series is convergent, but not absolutely convergent. Find a bound for the truncation error after 99 terms.

n 1

n=1

-11 1 11 - + - + ... +

2 3 4 n

Alternating Harmonic Series

n 1

n=1

n

-11 1 1 1 - + - + ... +

2 3 4 n

Since the terms are alternating in sign and decrease in absolute value

1 1 1 1 1 > > > ..., and since lim 0

Alt

2 3 4 n

By the ernating Series

n

n=1

1

n=1

.

1However, the series is the harmonic series which diverges.

n

So, the alternating harmonic series converges but is not absolutely convergent.

-1Test, converges

n

Truncation error after 99

100

1 terms is < u =

100

Rearranging Alternating Harmonic Series

The series of positive terms

1 1 1 1 + + + ... + diverges to

3 5 2n + 1While the series of negative terms

1 1 1 1 - - - - ... - diverges to -

2 4 6 2nSo, start by adding positive terms unti

l sum > 1, then add negative

terms until sum is less than -2, then add terms until sum > 3 etc so the sum

swings further in both directions and thus diverges.

To get the sum of , add positive terms unt il the partial sum is greater

than , then add negative terms until the sum is less than . Continue

indefinitely always getting closer to . Since the partial sums go to 0,

the sum gets converges to

.

Word of Caution

Although we can use the tests we have developed to find where a given power series converges, it does not tell us what function that power series is converging to. That is why it is so important to estimate the error.

Maclaurin Series of a Strange Function

21-

x

(n)

0, x = 0Let f(x) =

e , x 0

f (x) has derivatives of all orders at x = 0 and f (0) = 0 for all n

1. Construct the Maclaurin series for f.

2. For what values of x does this series converge?

3. Find all values of x for which the series actually converges to f(x).

Maclaurin Series of a Strange Function

21-

x

(n)

0, x = 0Let f(x) =

e , x 0

f (x) has derivatives of all orders at x = 0 and f (0) = 0 for all n

1. Construct the Maclaurin series for f.

2. For what values of x does this series converge?

The series converges to 0 for all values of x.

3. Find all values of x for which the series actually converges to f(x). The only place that this series actually converges to its f-value is at x = 0

n

n 0

x0 = 0

n!

Series DivergesIs lim an = 0?nth-Term Test

no

Geometric

Series TestIs Σ an = a + ar + ar2 + …? Converges to a/(1 - r) if |r| < 1.

Diverges if |r| ≥ 1.

p-series Test Does the series have the form

p

n 1

1

n ?

Series converges if p > 1Series diverges if p ≤ 1

yes

yes

Absolute

convergenceDoes Σ |an| converge? Apply 1 of the Comparison tests, IntegralTest, Ratio Test or nth-Root Test

Original series convergesyes

Alternating

Series TestIs Σ an = u1 – u2 + u3 - …?

Is there an integer N such thatun ≥ un+1 ≥… ?

Series converges if un →0.Otherwise series diverges.

no

no

no

no

Try partial sums

yes

yes