Do Now: Exploration 1 on p.492

Do Now: Exploration 1 on p.492

Consider and1

1n n

2

1

1n n

1. Show that the Ratio Test yields for both series.1L

For :1

1n n

1 1

lim1n

nL

n

lim

1n

nn

1

For :21

1n n

2

2

1 1lim

1n

nL

n

2

2lim1n

nn

1

Do Now: Exploration 1 on p.492

Consider and1

1n n

2

1

1n n

2. Use improper integrals to find the areas shaded in Figures9.13a and 9.13b for .

(a)1

1 dxx

1 x

1

1limk

kdx

x 1lim ln k

kx

lim ln ln1k

k

(b) 21

1 dxx

2

1lim

k

kx dx

1

1lim

k

kx

lim 1 1

kk

1

Do Now: Exploration 1 on p.492

Consider and1

1n n

2

1

1n n

3. Explain how Figure 9.14a shows that diverges,while Figure 9.14b shows that converges.

1 n21 n

Figure 9.14a shows that is greater than .1n 1

1 dxx

Since the integral diverges, so must the series.

Figure 9.14b shows that is less than .2

1n 21

11 dxx

Since the integral converges, so must the series.

Do Now: Exploration 1 on p.492

Consider and1

1n n

2

1

1n n

4. Explain how this proves the last part of the Ratio Test.

These two examples prove that L = 1 can be true foreither a divergent series or a convergent series. TheRatio Test itself is therefore inconclusive when L = 1.

ADDITIONAL CONVERGENCE

TESTSSection 9.5a

The Integral TestLet be a sequence of positive terms. Suppose that na

, where f is a continuous, positive, decreasing na f nfunction of x for all (N a positive integer). Then thex Nseries and the integral either bothnn N

a

N

f x dx

converge or both diverge.

For the graphical proof, we will let N = 1 for the sake ofsimplicity, but the illustration can be shifted horizontally toany value of N without affecting the logic of the proof.

The Integral Test

1 2 3 n n + 1na2a1a

y f x

(a)1 2 3 n – 1 n

na3a2a y f x

(b)

1a

(a) The sum provides an upper bound for1 2 na a a

1

1

nf x dx

(b) The sum provides a lower bound for2 3 na a a

1

nf x dx

Applying the Integral Test

Does converge?1

1n n n

The integral test applies because the function 1f xx x

is continuous, positive, and decreasing for x > 1.

Check the integral:1

1 dxx x

3 2

1lim

k

kx dx

1 2

1lim 2

k

kx

2lim 2k k

2Since the integral converges, so must the series.

(side note: they do not have to converge to the same value)

Harmonic Series and p-seriesAny series of the form (p a real constant) is 1

1 pn

n

called a p-series. The p-series test:1. Use the Integral Test to prove that convergesif p > 1.

11 p

nn

1

1p dx

x

1

1limk

pkdx

x

1

1

lim1

kp

k

xp

1

1 1lim 11 pk p k

1 0 1

1 p

11p

The series converges.

Harmonic Series and p-seriesAny series of the form (p a real constant) is 1

1 pn

n

called a p-series. The p-series test:2. Use the Integral Test to prove that divergesif p < 1.

11 p

nn

1

1p dx

x

1

1limk

pkdx

x

1

1

lim1

kp

k

xp

11lim 11

p

kk

p

The series diverges.

If 0 < p < 1:

0p(If , the series diverges by the nth-Term Test)

Harmonic Series and p-seriesAny series of the form (p a real constant) is 1

1 pn

n

called a p-series. The p-series test:3. Use the Integral Test to prove that divergesif p = 1.

11 p

nn

1

1p dx

x

1

1limk

kdx

x 1lim ln k

kx

lim ln ln1k

k

The series diverges.

This last divergent series is called the harmonic series.

The Limit Comparison Test (LCT)Suppose that and for all0na 0nb n N

(N a positive integer).

1. If then andlim ,n

nn

a cb 0 ,c na nb

both converge or both diverge.

2. If and converges, thenlim 0n

nn

ab nanb

converges.

3. If and diverges, thenlim n

nn

ab nanb

diverges.

Using the LCTDetermine whether the given series converge or diverge.

21

3 5 7 9 2 14 9 16 25 1n

nn

For n large, behaves like 22 1

1n

n

2

2 2nn n

Compare the given series to and try the LCT: 1 n

lim n

nn

ab

22 1 1lim

1n

n nn

22 1lim

11n

n nn

Using the LCTDetermine whether the given series converge or diverge.

21

3 5 7 9 2 14 9 16 25 1n

nn

Since the limit is positive and diverges, 1 n 22 1lim

11n

n nn

2

2

2lim2 1n

n nn n

2

21

2 11n

nn

also diverges.

Using the LCTDetermine whether the given series converge or diverge.

1

1 1 1 1 11 3 7 15 2 1n

n

For n large, behaves like 1 2 1n 1 2n

Compare the given series to : 1 2nlim n

nn

ab

1 2lim2 1 1

n

nn

2lim2 1

n

nn

1lim

1 1 2nn

1

Since 1 2nconverges (thisis a geometric

series withr = 1/2), thegiven series

also converges.

Using the LCTDetermine whether the given series converge or diverge.

32

8 11 14 17 3 24 21 56 115 2n

nn n

For n large, the series behaves like 23 n: 21 n

lim n

nn

ab

Compare to

2

3

3 2lim2 1n

n nn n

3 2

3

3 2lim2n

n nn n

3

Since converges by the p-Series Test, 21 nthe given series also converges (by the LCT).

More Guided PracticeDetermine whether the series converges or diverges. Theremay be more than on correct way to determine convergenceor divergence of a given series.

1

3n n

1 2

1

13n n

The series diverges bythe p-Series Test

1

1ln 3 n

n

The series converges,since it is geometric with

1 0.910ln 3

r

More Guided PracticeDetermine whether the series converges or diverges. Theremay be more than on correct way to determine convergenceor divergence of a given series.

1

12 1n n

The Integral Test:

1

12 1

dxx

1

1lim2 1

k

kdx

x

1

1lim ln 2 12

k

kx

1lim ln 2 1

2kk

Since the integral diverges, the series diverges as well.