Section 11.5 – Testing for Convergence at Endpoints.
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Transcript of Section 11.5 – Testing for Convergence at Endpoints.
Section 11.5 – Testing for Convergence at Endpoints
Common Series to be used….
Harmonic Series - n 1
1 1 1 11 ...
n 2 3 4
DIVERGES
p-Seriesp p p p
n 1
1 1 1 11 ...
n 2 3 4
Converges if p > 1Diverges if p < 1
Comparison Test for Convergence
n n n nconverges conva if c and a c for alle ges nr
Comparison Test for Divergence
n n n ndiverges divea if d and a drges for all n
3n 1
n 1
n 3
nnlima 0
3
n 1
n 3
acts like 2
1as n
n
3n 1
n 1
n 3
converges by the comparison test
to the p-series with p = 2.
Makes terms smaller
Makes terms smaller
2
2n 1
n 1
n 3n 2
n
nlima 1
2
n 2nn 1
n 1Since lima 1,
n 3n 2
diverges by the nth term test
for divergence.
2
3n 1
6n 1
2n 1
n
nlima 0
2
3
6n 1
2n 1
acts like3
as nn
2
3n 1
6n 1
2n 1
diverges by the comparison test to
the p-series with p = 1.
n nIf a converges, then a converges
n
n
A series a is said to CONVERGE ABSOLUTELY
if a converges
ABSOLUTE CONVERGENCE
2n 1
1converges absolutely
n 1
by comparison to p-series with p=2>1.
2n 1
1Test for convergence.
n 2n 1
n 1
2n 1
11
n
2n 1
1Since is a p-series with p = 2>1, it converges.
n
n 1
2n 1
11 converges absolutely.
n
2n 1
1 n
n
2n 1
1Since converges (p-series with p = 2>1)
n
n 1
1 and diverges (harmonic series)
n
2n 1
1 ndiverges
n
n 1
cosn
n n
n 1
cosnconverges absolutely
n n
3
by comparison to p-series with p= >1.2
n
nn 1
n n1 diverges since lim 1 0
n 1 n 1
n
n 1
5
n
n 1
5 converges
-1
since infinite geometric series with r = 5
n
n 1
n1
n 1
n n 1
n n nn=1 n=1
If a 0, then an alternating series -1 a or -1 a
Alternating Series Test
converges if both of the following conditions are satisfied:
n
n
n n+1 n
1. lima 0
2. a is a decreasing sequence for all n, or a a
If a series converges but the series of absolute values diverges,We say the series converges conditionally.
CONDITIONAL CONVERGENCE – (AST)
Determine whether the series is converges conditionally, converges absolutely, or diverges.
n 1
cos n
n
n
11. lim 0
n
1 12.
n 1 n
The series converges conditionally
n 1
cosndoes not converge absolutely
n by comparison test to the p-series with p=1/2<1
Determine whether the series is converges conditionally, converges absolutely, or diverges.
n 1 5
n 1
1 n
n 1 5
n 1
1 n
by comparison test t
converges absol
o the p-series
utely
with p=5>1
Determine whether the series is converges conditionally, converges absolutely, or diverges.
n 1
n 1
n1
n 1
n
n1. lim 1
n 1
The series diverges by the nth term test for divergence
n 1 n n 1
nn 1 n 1 n 1
Which of the following series are conditionally convergent?
1 cosn 1I. 1 II. 1 III. 1
2n 1 3 nA. I only B. II only C. I and II only D. I and III only E. I, II, III
n 1
n 0n 1
1 1I. 1 1. lim 0
2n 1 2TRUE
n 1
1 12.
2 n 1 1 2n 1TRUE
n
n nn 0n 1
cosn cosnII. 1 1. lim 0
3 3TRUE 1
n 1
n 0n 1
1 1III. 1 1. lim 0
n n1 1
2.n
TRUE
TRU1
En
2 2 2
n
n
2
Which of the following series are convergent?
1 1 1I. 1 ... ...
2 3 n
11 1II. 1 ... ...
2 3 n
8 2III. 2 1 ... ...
9 nA. I only B. III only C. I and II only D. II and III only E. I, II, III
I. p-series with p CON= 2 VE >1 RGES
II. alt. harmonic CONser VEies RGES
2n 1 2
2 nn n
2 n nl
III. ratio
im lim 2 22 n 1n
DIVERG
1
E test S
Find the interval of convergence for n
n 1
x
n
n 1
nn
x nlim 1
n 1 x
n
nx lim 1
n 1
x 1
If x = -1, n
n 1
1
n
converges, alternating series test
If x = 1, n
n 1
1
n
diverges, harmonic series
Interval of convergence [-1, 1)
1 x 1 ? ?
Find the interval of convergence for n
nn 0
nx 3
4
n 1 n
n 1 nn
n 1 x 3 4lim 1
4 n x 3
n
n 1x 3 lim 1
4n
x 3 4
If x = -7, n
nn 1
n
4
4
diverges, nth term test
If x = 1, n
nn 1 n 14
4nn
diverges, nth term test
Interval of convergence (-7, 1)
n
n 1
1 n
7 x 1 ? ?
Find the interval of convergence for n n
n 0
2 x
n!
n 1 n 1
n nn
2 x n!lim 1
n 1 ! 2 x
n
2x lim 1
n 1
0 1
Interval of convergence ,
Find the interval of convergence for
n 1 n 1
n nn
1 x 3 n 1lim 1
n 2 1 x 3
n
n 1x 3 lim 1
n 2
x 3 1
If x = 2, n n
n 0
1
n
1
1
diverges, harmonic series
If x = 4, n n n
n 1 n 1
1 1
n 1 n 1
1
converges, alternating series test
Interval of convergence (2, 4]
n n
0
1 x 3
n 1
2 x 4 ? ?
5
1 1a
2 5 10
5 1
5
1 1a
3 729
5 3
1 1a
1255
Find an upper bound for the error if the sum of the first fourterms is used as an approximation to the sum of the series.
n 1
n 1
114.
2n
n 1
n 1
116.
3
n 1
3n 1
118.
n
n 1a 0.005
1 1
2 n 1 200
2 n 1 200
n 99
n 1a 0.005
n 2
1 1
2003
n 2003
9
n 3
Find the smallest value of n for which the nth partial sum approximates the sum of the series within 0.005.
n 1
n 1
114.
2n
n 1
n 1
116.
3
n 1
3n 1
118.
n
n 1a 0.005
3
1 1
200n 1
3n 1 200
n 5