Section 8.6: Alternating Series
-
An Alternating Series is of the form
 (1)
k
ak
or
(with ak >0)
 (1)
k 1
ak

 (1)
k 1
k 1
1
1 1 1 1 1
 1      ...
k
2 3 4 5 6
1
1
1
 0.5
2
1 1
1 
 .83333333...
2 3
1 1 1
1  
 0.583333....
2 3 4
1 1 1 1
1   
 .7833333....
2 3 4 5
1 1 1 1
1
1   

 .6166666....
2 3 4 5
6
1
0.8333...
0.78333...
0.61666...
0.58333...
0.5
Alternating Series test
If
k
(

1)
ak

is an alternating series,
ak  is a decreasing sequence, and
lim ak  0
k 
Then
 (1)
k
ak
converges
1
  1 k
k
Converges by the
Alternating Series Test
3k
  1 4k  1
k
2
k
  1 k 3  4
k
Diverges by the
Divergence Test
Converges by the
Alternating Series Test
Definition
a
k
if
If
converges absolutely

| ak | converges
 a converges , but not absolutely,
k
then it converges conditionally.
1
  1 k 2
k
1
  1 k
k
Converges absolutely
Converges conditionally
cos k
 k2

cos k
2
k
So 
cos k
2
k
cos k
1
 2
2
k
k
converges
by Direct Comparison Test
cos k
 k2
Thus
Converges Absolutely
Generalized Ratio Test
Recall the Ratio test required ak  0
ak 1
but we could just check lim
k 
ak
and see if
 a converges absolutely.
k
Error Estimates
Notice if  (1) ak satisfies the
Alternating Series Test,
k
then it is a
“1 step forward, half step back”
kind of sum.
One term pushes sum up above the limit,
the next pushes it below the limit..
So the difference between the partial
sum and the true sum is less than
the next term!

 (1)
k 1
k 1
1
1 1 1 1 1
 1      ...
k
2 3 4 5 6
1
1
1
 0.5
2
1 1
1 
 .83333333...
2 3
1 1 1
1  
 0.583333....
2 3 4
1 1 1 1
1   
 .7833333....
2 3 4 5
1 1 1 1
1
1   

 .6166666....
2 3 4 5
6
1
0.8333...
0.78333...
0.61666...
0.58333...
0.5
Rearranging terms
Its probably not too surprising that
you can rearrange the terms
in an absolutely convergent series
and still get the same sum,
but a conditionally convergent sum
can be rearranged
to sum to anything you want.
Proof:
Pick any number L you’d like the
new series to converge to.
The positive terms and negative terms
both converge to 0 as sequences,
but the sum of each diverges.
Re-arrange them from largest to smallest.
Add up positive terms until you get past L.
Subtract negative pieces until
you get below L.
Repeat…
Since terms 0 the “over shooting” 0