Download - 9.4 Part 1 Convergence of a Series
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
