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11.29
Definition (Alternating series.)
Series of the form

or

where

for all

are called
alternating series.
11.30
Theorem (Alternating series test.)
Let
be a decreasing sequence of positive numbers such that
. Then
is summable. Moreover,
and
for all
.
Proof: Let
. For all
,
and
Thus
is decreasing and
is increasing. Also, for all
,
so
is bounded below by
, and
so
is bounded above by
.
It follows that there exist real numbers
and
such that
Now
so
.
It follows from the next lemma that
; i.e.,
Since for all
we have
and since
Thus, in all cases,
; i.e.,
approximates
with an error of no more than
.
11.31
Lemma.
Let
be a real sequence and let
. Suppose
and
. Then
.
Proof: Let
be a precision function for
and let
be a precision
function for
. For all
, define
I claim
is a precision function for
, and hence
.
Let
.
- Case 1:
is even. Suppose
is even.
Say
where
. Then
- Case 2:
is odd.
Suppose
is odd. Say
where
.
Then
Hence, in all cases,
11.32
Remark.
The alternating series test has obvious generalizations for series such as
and we will use these generalizations.
11.33
Example.
If

, then
are decreasing positive null sequences, so
are summable; i.e.,
(These are the sequences we called

and

in example
10.3.)
Also,
,
with an error smaller than
. My calculator says
and
11.34
Entertainment.
Since

is a decreasing positive null
sequence for

, it follows that

converges for

. We will now explicitly
calculate the limit of this series using a few ideas that are not justified by results
proved in this course. We know that for all

, and all

,
Hence, for all

,
i.e.,
Thus
Hence
for all

.
If we can show that
is a null
sequence, it follows that
or in other words,
 |
(11.35) |
I claim

is a null sequence for

and hence (
11.35) holds for

. In particular,
First suppose

, then

for

, so
Since

is a null sequence for

, it follows from the comparison test that

is a null sequence for

.
Now suppose

. Then
so

and
If

, then

is a
null sequence, so

is a null
sequence.
11.36
Entertainment.
By starting with the formula
for all

and using the ideas from the last example, show that
![\begin{displaymath}
\sum_{j=0}^\infty {{(-1)^jx^{2j+1}}\over {(2j+1)}}=\arctan(x)\mbox{ for all }x\in[-1,1].
\end{displaymath}](img1599.gif) |
(11.37) |
Conclude that
Next: 11.4 Absolute Convergence
Up: 11. Infinite Series
Previous: 11.2 Convergence Tests
  Index