do now: exploration 1 on p.492
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Do Now: Exploration 1 on p.492. Consider and. 1. Show that the Ratio Test yields for both series. For :. For :. Do Now: Exploration 1 on p.492. Consider and. 2. Use improper integrals to find the areas shaded in Figures - PowerPoint PPT PresentationTRANSCRIPT
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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
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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
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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.
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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.
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ADDITIONAL CONVERGENCE
TESTSSection 9.5a
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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.
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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
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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)
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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.
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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)
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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.
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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.
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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
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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.
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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.
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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).
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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
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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.