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