Series with Positive terms: tests for
Convergence, Pt. 1The comparison test,
the limit comparison test, and the integral test.
Comparing series. . .Consider two series ,
with for all k.1 1
and k kk k
a b
kk ba 0
In this presentation, we will base all our series at 1, but similar results apply if they start at 0 or elsewhere.
Comparing series. . .Consider two series ,
with for all k.
How are these related in terms of convergence or divergence?
1 1
and k kk k
a b
kk ba 0
Note that:
1 1
1 2 1 2
1 2 3 1 2 3
1 2 3 4 1 2 3 4
And so on
a ba a b ba a a b b ba a a a b b b b
What does this tell us?
Comparing series. . .Consider two series ,
with for all k.
and k ka b
kk ba 0
Note that:
1 1
1 2 1 2
1 2 3 1 2 3
1 2 3 4 1 2 3 4
And so on
a ba a b ba a a b b ba a a a b b b b
What does this tell us?
Where does the fact that the terms are non-negative come in?
Series with positive terms. . .
1x
Since for all positive integers k. Then
0 kx
So the sequence of partial sums is . . . 1
n
n kk
s a
Non-decreasing Bounded above Geometric
1 2x x 1 2 3x x x 1 2 3 4x x x x
Back to our previous scenario. . . Consider two series ,
with for all k.1 1
and k kk k
a b
0 k ka b
Suppose that the series converges 1
kk
b
Suppose that the series converges
0 0 0
Note that for all positive integers ,
n n
k k kk k k
n
a b b
1k
k
b
Non-decreasing Bounded above Geometric
So the sequence of partial sums is . . . 1
n
n kk
s a
A variant of a familiar theorem
Suppose that the sequence is non-decreasing and bounded above by a number A. That is, . . .
ks
1 1 2 1 2 3a a a a a a A
Then the series converges to some value that is smaller than or equal to A.
1k
k
a
Theorem 3 on page 553 of OZ
Suppose that the series diverges
0 0
n n
k kk k
a b
1k
k
a
Non-decreasing Bounded below Unbounded
So the sequence of partial sums is . . . 1
n
n kk
s b
For all n we still have
This gives us. . . The Comparison Test: Suppose we have two series , with for all positive integers k.
If converges, so does , and
If diverges, so does .
1 1
and k kk k
a b
1k
k
b
1k
k
a
1k
k
b
1k
k
a
kk ba 0
A related test. . .
There is a test that is closely related to the comparison test, but is generally easier to apply. . . It is called the
Limit Comparison Test
This test is not in the book!
(One case of…) The Limit Comparison Test
Limit Comparison Test: Consider two series
with , each with positive terms.
If , then
are either both convergent or both divergent.
1 1
and k kk k
a b
n
n
ba
nlim
01 1
and k kk k
a b
Why does this work?
(Hand waving) Answer:
Because if
Then for “large” n, ak t bk. This means that “in the long run”
0 0
and have the same convergence behavior.n n
k kk k
a t b
tba
n n
n
lim
The Integral Test
y = a(x)
Now we add some enlightening pieces to our diagram….
Suppose that we have a sequence {ak} and we associate it with a continuous function y = a(x), as we did a few days ago. . .
The Integral Test
y = a(x)
Look at the graph. . .What do you see?
Suppose that we have a sequence {ak} and we associate it with a continuous function y = a(x), as we did a few days ago. . .
y = a(x)
The Integral Test
1a2a
3a4a
6a5a7a
If the integral
so does the series.
11
( ) ( )k
a x dx a k
So
converges
diverges
The Integral Test
Now look at this graph. . .What do you see?
y = a(x)
y = a(x)
The Integral Test
1a
2a3a
4a6a5a
7a
If the integral
so does the series.
12
( ) ( )k
a k a x dx
So
converges
diverges8a
Why 2?
The Integral TestThe Integral Test: Suppose for all x 1, the function a(x) is continuous,
positive, and decreasing. Consider the series and the integral .
If the integral converges, then so does the series.If the integral diverges, then so does the series.
1
( )k
a k
1( )a x dx
The Integral TestThe Integral Test: Suppose for all x 1, the function a(x) is continuous,
positive, and decreasing. Consider the series and the integral .
If the integral converges, then so does the series.If the integral diverges, then so does the series.
1
( )k
a k
1( )a x dx
Where do “positive and decreasing”
come in?
