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9.4 Part 1 Convergence of a Series. n th term test for divergence. diverges if fails to exist or is not zero. The first requirement of convergence is that the terms must approach zero. Note that this can prove that a series diverges , but can not prove that a series converges. - PowerPoint PPT Presentation
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9.4 Part 1 Convergence of a Series
The first requirement of convergence is that the terms must approach zero.
nth term test for divergence
1 nna
diverges if fails to exist or is not zero.lim nna
Note that this can prove that a series diverges, but can not prove that a series converges.
The first requirement of convergence is that the terms must approach zero.
nth term test for divergence
1 nna
diverges if fails to exist or is not zero.lim nna
1
1
n
ne
n
ne
1
lim 01
So eventually, as n ∞, the sum goes to 1 + 1 + 1 + 1…
So the series diverges
So eventually, as n ∞, the sum goes to 0 + 0 + 0 + 0…
The first requirement of convergence is that the terms must approach zero.
nth term test for divergence
1 nna
diverges if fails to exist or is not zero.lim nna
1n
ne
n
nelim
nn e
1lim
So the series may converge
0
This series converges.
So this series must also converge.
Direct Comparison Test For non-negative series:
If every term of a series is less than the corresponding term of a convergent series, then both series converge.
0 4
3
n
n
0 41
3
nn
n
is a convergent geometric series
But what about…
n
n
n
n
4
3
41
3
for all integers n > 0
So by the Direct Comparison Test, the series converges
Direct Comparison Test For non-negative series:
If every term of a series is greater than the corresponding term of a divergent series, then both series diverge.
So this series must also diverge.
This series diverges.
0 3
4
n
n
0 3
41
nn
n
is a divergent geometric series
But what about…
n
n
n
n
3
4
3
41
for all integers n > 0
So by the Direct Comparison Test, the series diverges
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.
Notice also that the terms go to 0 yet it still diverges
01
lim nn
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 :
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
Another series for which it is easy to find the sum is the telescoping series.
Ex. 6: 1
1
1n n n
Using partial fractions:
1 A 0 A B
0 1 B
1 B
1
1 1
1n n n
1 1 1 11 1
2 3 3 42
3
11
4S
11
1nS n
lim 1nnS
1
11
A B
n n nn
1 1A n Bn
1 An A Bn
Telescoping Series
11
n nn
b b
converges to 1b
Another series for which it is easy to find the sum is the telescoping series.
Ex. 6: 1
1
1n n n
1
1 1
1n n n
1 1 1 11 1
2 3 3 42
3
11
4S
11
1nS n
lim 1nnS
