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Lecture 20

MAT187H1 Lecture 20: Divergence Integral Tests 9.5 Ratio Test 9.6 Alternating Series and Absolute and Conditional ConvergencePremium


Department
Mathematics
Course Code
MAT136H1
Professor
Anthony Lam
Lecture
20

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MAT187H1Lecture209.4Divergence&IntegralTests,9.5RatioTest,9.6Alternating
SeriesandAbsoluteandConditionalConvergence
ExampleApproximate using andboundsonremainder.Howmanytermsare
6
π2Sn= ∑
k=1
1
k2
neededforE≤10
3
L
n
=S+ U
n
=S+
n+1 x2
dx =1
n+1
nx2
dx =n
1
nSn= ∑
k=1
1
k2L
n
=S+ U
n
=S+
n+1 x2
dx =1
n+1
nx2
dx =n
1
1 1 1+½=3/2 1+1=2
2 1+¼=5/4 5/4+⅓=19/12 5/4+½=7/4
3 5/4+1/9=49/36 49/36+¼=58/36 49/36+⅓=61/36
Note: .644934
k=1
1
k2=6
π2= 1
ForE=10
3
:R
n
=SS
n
≤10
3
 n>1000≤10
nx2
dx =n
1−3
Note: ..
k=1
1
k4= 1 + 1
24+1
34+ . = π4
90
9.5RatioTest(andRootandComparisonTests)
RatioTestforPositiveTermSeries
Define: ρ lim
n→∞ an
an+1
Ifρ<1,thenΣa
n
converges
Ifρ>1,orρ=∞,thenΣa
n
diverges
Ifρ=1,testinconclusive
(inessence,forρ<1,thena
n+1
>ρa
n
,behavelikeageometricseries)
RootTest
Define ρ lim
n→∞
nan= lim
n→∞ an
n
1
Ifρ<1,thenΣa
n
converges
Ifρ>1,thenΣa
n
diverges
Ifρ=1,testinconclusive
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