Series with Positive
terms: tests for
Convergence, Pt. 1
The comparison test,
the limit comparison test,
and the integral test.
Comparing series. . .
Consider two series ,

a
k 1
k

and
b
k 1
k
with 0  ak  bk for all k.
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 ,

a
k 1
k
Note that:

and
b
k 1
k
with 0  ak  bk for all k.
How are these related in
terms of convergence or
divergence?
a1  b1
a1  a2  b1  b2
a1  a2  a3  b1  b2  b3
a1  a2  a3  a4  b1  b2  b3  b4
And so on
What does this tell us?
Comparing series. . .
Consider two series ,
a
k
and  bk
with 0  ak  bk for all k.
Where does the fact
that the terms are
non-negative come
in?
Note that:
a1  b1
a1  a2  b1  b2
a1  a2  a3  b1  b2  b3
a1  a2  a3  a4  b1  b2  b3  b4
And so on
What does this tell us?
Series with positive terms. . .
Since
x1 
0  xk for all positive integers k. Then
x1  x2 
x1  x2  x3  x1  x2  x3  x4 
n 
So the sequence of partial sums sn    ak  is . . .
 k 1 
Non-decreasing
Bounded above
Geometric
Back to our previous scenario. . .
Consider two series ,

a
k 1
k

and
b
k 1
k
with 0  ak  bk for all k.

Suppose that the series  bk converges
k 1

Suppose that the series  bk converges
k 1
Note that for all positive integers n,
n

n
a  b  b
k 0
k
k 0
k
k 0
k
n 
So the sequence of partial sums sn    ak  is . . .
 k 1 
Non-decreasing
Bounded above
Geometric
A variant of a familiar theorem
Suppose that the sequence sk  is non-decreasing
and bounded above by a number A. That is, . . .
Theorem 3 on
a  apage
 a553of OZa  a  a 
 A
1
1
2
1
2
3

Then the series
a
k 1
k
converges to some value that
is smaller than or equal to A.

Suppose that the series  ak diverges
k 1
n
For all n we still have
n
a  b
k 0
k
k 0
k
n 
So the sequence of partial sums sn    bk  is . . .
 k 1 
Non-decreasing
Bounded below
Unbounded
This gives us. . .
The Comparison Test:

Suppose we have two series ,  ak and
k 1
0  ak  bk for all positive integers k.


If
k
converges, so does
If
a
k 1
a
k 1
k
k 1
, and



b

b
k 1

k
diverges, so does
b
k 1
k
.
k
with
A related test. .This
. test is not in
the
book!
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
(One case of…)
The Limit Comparison Test
Limit Comparison Test: Consider two series

with  ak and
k 1

If
lim

b
k 1
k
, each with positive terms.
an
0
 , then
n   bn

a
k 1
k

and  bk
k 1
are either both convergent or both divergent.
Why does this work?
(Hand waving)

Answer:
Because if
lim
an
t
n   bn
Then for “large” n, ak  t bk. This means that “in the
long run”
n
a
k 0
k
n
and t  bk have the same convergence behavior.
k 0
The Integral Test
y = a(x)
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. . .
Now we add some
enlightening pieces to our
diagram….
The Integral Test
y = a(x)
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. . .
Look at the graph. . .
What do you see?
The Integral Test
So


1

a( x) dx   a(k )
y = a(x)
k 1
converges
a1
If the integral
a2
a3 a
4 a
5 a6 a
7
diverges
so does the series.
The Integral Test
y = a(x)
Now look at this graph. . .
What do you see?
The Integral Test

So
 a(k )  
k 2
y = a(x)
Why 2?
a1

1
a( x) dx
converges
If the integral
a2
a3 a
4 a
5 a6 a
a8
7
diverges
so does the series.
The Integral Test
The Integral Test:
Suppose for all x  1, the function a(x) is continuous,
positive, and decreasing. Consider the series

 a(k )
k 1
and the integral


1
a( x) dx .
If the integral converges, then so does the series.
If the integral diverges, then so does the series.
The Integral Test
The Integral Test:
Suppose for all x  1, the function a(x) is continuous,
positive, and decreasing. Consider the series

 a(k )


and Where
the integral
do “positive
a( x) dx .
1
and decreasing”
k 1
come
If the integral converges, then
so in?
does the series.
If the integral diverges, then so does the series.