Alternating Series
The Alternating Series Test
Suppose that a weight from a spring is released. Let a1
be the distance that the spring drops on the first bounce. Let
a2 be the amount the weight travels up the first time. Let
a3 be the amount the weight travels on the way down for
the second trip. Let a4 be the amount that the weight travels
on the way up for the second trip, etc.
Then eventually the weight will come to rest somewhere in the middle.
This leads us to
Theorem: The
Alternating Series Test
Let
an > 0
for all n
and suppose that the following two conditions
hold:
-
{ an }
is a decreasing sequence for large
n.
-
Then the corresponding series
and
converge. |
Proof:
We will prove the theorem for the second given series. This is enough,
since the first can be obtained from the second just by multiplying by
-1.
We look at the series as adding two at a time and then adding them all together.
s2n = (a1 - a2) + (a3 -
a4) + ...+ (a2n-1 - a2n) > 0
which shows that this is bounded below by 0.
Now single out the first term and then add the rest two at a time
s2n = a1 - (a2 - a3) -
(a4 - a5) - ...- (a2n-2 -
a2n-1) - a2n
= a1 -
[(a2 - a3) +
(a4 - a5) + ...+ (a2n-2 -
a2n-1) + a2n]
This second equation subtracts a positive number from the first term.
Hence
s2n < a1
which shows that the sequence is bounded above by a1.
Notice that s2n is monotonic since each
difference is positive. Therefore s2n is bounded and monotonic and thus converges.
Since the an tend toward zero as n
tends towards infinity, we have
The limit of the
partial sums exist and hence the series converges.
Example
converges by the alternating series test, since the
and
1
1
>
n
n + 1
Exercises:
Determine whether the following converge:
-
-
The Remainder Theorem
Consider the spring example again. The weight will always be between
the two previous positions. Hence we have
The Remainder Theorem
Let
then
|L - sn| < an + 1
|
This says that the error in using n terms to
approximate an alternating series is always less then the n
+ 1st term.
Example
Use a calculator to determine
With an error of less than .01.
Solution:
We have
Error < .01
so choose n such that
1
< .01
n
Here, n = 101 will work. Then use your calculator to get
0.70.
Absolute and Conditional Convergence
Example:
The alternating harmonic series is conditionally convergent
since we saw before that it converges by the alternating series test but its
absolute value (the harmonic series) diverges.
The series
is absolutely convergent since the series of the absolute value of its terms is
a P-series with p = 2, hence converges.
The Rearrangement Theorem
The Rearrangement Theorem
Let
be a conditionally convergent series and
let k be a real number. Then there exists a rearrangement of the terms
so that you add them up and end up with k. As strange as it may seem, addition
is not commutative for conditionally convergent series. On the other
hand for absolutely convergent series any rearrangement produces the same
limit. |
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