9.5 testing convergence at endpoints greg kelly, hanford high school, richland, washington the...

9.5 Testing Convergence at Endpoints

Greg Kelly, Hanford High School, Richland, WashingtonThe original Hanford High School, Hanford, Washington

Remember:

The series converges if .1L

The series diverges if .1L

The test is inconclusive if .1L

The Ratio Test:

If is a series with positive terms andna 1lim n

nn

aL

a

then:

This section in the book presents several other tests or techniques to test for convergence, and discusses some specific convergent and divergent series.

The series converges if .1L

The series diverges if .1L

The test is inconclusive if .1L

Nth Root Test:

If is a series with positive terms andna lim nnna L

then:

Note that the rules are the same as for the Ratio Test.

example:2

1 2nn

n

2

2n

n

n 2

2

n n

2lim n

nn

2

lim n

nn

?

lim n

nn

1

lim n

nn

1lim ln nn ne

1lim lnn

nne

lnlimn

n

ne

1

lim1n

n

e

0e

1

Indeterminate, so we use L’Hôpital’s Rule

formula #104

formula #103

example:2

1 2nn

n

2

2n

n

n 2

2

n n

2

lim2

n

n

n

2lim n

nn

2

lim n

nn

21 1

1

2 it converges

?

another example:2

1

2n

n n

2

2nn

n 2

2n n

2

2lim

nn n

2

1 it diverges2

Remember that when we first studied integrals, we used a summation of rectangles to approximate the area under a curve:

This leads to:

The Integral Test

If is a positive sequence and where

is a continuous, positive decreasing function, then:

na na f n f n

and both converge or both diverge.nn N

a

Nf dxx

Example 1: Does converge?1

1

n n n

1

1 dx

x x

3

2

1lim

b

bx dx

1

21

lim 2

b

b x

22lim

b b

2

Since the integral converges, the series must converge.

(but not necessarily to 2.)

p-series Test

1

1 1 1 1

1 2 3p p p pn n

converges if , diverges if .1p 1p

We could show this with the integral test.

If this test seems backward after the ratio and nth root

tests, remember that larger values of p would make the

denominators increase faster and the terms decrease

faster.

the harmonic series:

1

1 1 1 1 1

1 2 3 4n n

diverges.

(It is a p-series with p=1.)

It diverges very slowly, but it diverges.

Because the p-series is so easy to evaluate, we use it to compare to other series.

Limit Comparison Test

If and for all (N a positive integer)0na 0nb n N

If , then both and

converge or both diverge.

lim 0n

nn

ac c

b na nb

If , then converges if converges.lim 0n

nn

a

b na nb

If , then diverges if diverges.lim n

nn

a

b na nb

Example 3a:

21

3 5 7 9 2 1

4 9 16 25 1n

n

n

When n is large, the function behaves like:2

2 2n

n n

2 1

n n lim n

nn

a

b

2

2 1

1lim1n

n

n

n

2

2 1lim1n

nn

n

2

2

2lim

2 1n

n n

n n

2

Since diverges, the

series diverges.

1

nharmonic series

Example 3b:

1

1 1 1 1 1

1 3 7 15 2 1nn

When n is large, the function behaves like:1

2n

lim n

nn

a

b

12 1lim

12

n

n

n

2lim

2 1

n

nn

1

Since converges, the series converges.1

2ngeometric series

Alternating Series The signs of the terms alternate.

Good news!

example: 1

1

1 1 1 1 1 1 11

1 2 3 4 5 6n

n n

This series converges (by the Alternating Series Test.)

If the absolute values of the terms

approach zero, then an alternating

series will always converge!

Alternating Series Test

This series is convergent, but not absolutely convergent.

Therefore we say that it is conditionally convergent.

Since each term of a convergent alternating series moves the partial sum a little closer to the limit:

Alternating Series Estimation Theorem

For a convergent alternating series, the truncation error is less than the first missing term, and is the same sign as that term.

This is a good tool to remember, because it is easier than the LaGrange Error Bound.

There is a flow chart on page 505 that might be helpful for deciding in what order to do which test. Mostly this just takes practice.

To do summations on the TI-89:

5

1

18

2

n

n

becomes ^ , ,1,5)8*(1/ 2 n n 31

4

1

18

2

n

n

becomes ^ , ,1, )8*(1/ 2 n n 8

F3 4

To graph the partial sums, we can use sequence mode.

MODE Graph……. 4 ENTER

Y= u1 ( 8*( 3/ 4) ^ , ,1, )k k n WINDOW

ENTER

GRAPH

To graph the partial sums, we can use sequence mode.

MODE Graph……. 4 ENTER

Y=

WINDOW

ENTER

GRAPH

Table

u1 ( 8*( 3/ 4) ^ , ,1, )k k n

To graph the partial sums, we can use sequence mode.

MODE Graph……. 4 ENTER

Y=

WINDOW

ENTER

GRAPH

Table

u1 ( 8*( 3/ 4) ^ , ,1, )k k n