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Definition

An alternating series is a series containing terms with alternating signs. Here are two examples:


 ∞
---     n-1
\   (-1)          1   1   1
/   ------- = 1 - - + - - - + ...
---    n          2   3   4
n=1

 ∞
---     n
\   (-1) n     1   2   3   4
/   ------ = - - + - - - + - + ...
---  n + 1     2   3   4   5
n=1

Looking at our examples we notice that the n-th term of an alternating series can be described in two ways:

an=(-1)n-1bn     or     an=(-1)nbn

Where bn is a positive number. More generally, bn = |an|.

Alternating Series Test

As is with all series mathematics, the fundamental question we must ask is, "Does this series converge?". With alternating series, mathematicians have come up with a special test to handle the case, dubiously named the Alternating Series Test.

If the alternating series

 ∞
---     n-1
\   (-1)   b  = b  - b  + b  - b  + ...   b  > 0
/           n    1    2    3    4          n
---
n=1

satisfies

1.) bn+1bn for all n
2.) The limit of bn as n approaches infinity equals zero.

Then the series converges.

Before giving a proof for this series test, I want to draw a picture that almost makes the proof obsolete. First we plot s1 = b1 on a number line, then find s2 by subtracting b2, etc, etc. We find that since bn approaches zero, the partial sums oscillate back and forth until the series converges to a sum. The even partial sums s2n are increasing and the odd decrease, and therefore it is plausible that both converge to some number s.


                         +b(1)
    |----------------------------------------->|
    |                       -b(2)              |
    |         |<-------------------------------|
    |         |       +b(3)                    |
    |         |------------------------>|      |
    |         |             -b(4)       |      |
    |         |       |<----------------|      |
    |         |       |                 |      |
----|---------|-------|---------|-------|------|---------
    0         s       s         s        s      s
               2       4                 3      1

Proof

s2 = b1 - b2 ≥ 0           since b2b1

s4 = s2 + (b3 - b4) ≥ s2    since b4b3

In general,         s2n = s2n-2 + (b2n-1 - b2n) ≥ s2n-2       since b2nb2n-1

Thus      0 ≤ s2s4s6 ≤ ... ≤ s2n ≤ ...

But we can also write:
s2n = b1 - (b2 - b3) - (s4 - s5) - ... - (b2n-2 - b2n-1) - b2n

Since every term in parentheses is positive, s2nb1 for all n. Therefore the sequence {s2n} or the even partial sums is increasing and bounded above by b1. Thus, by the Monotonic Sequence Theorem which says "Every bounded, monotonic sequence is convergent," our series must have a sum.

Well known Alternating Series

S(x) = sin(x)

        ∞
       ---           (2n+1)
S(x) = \       n    x
       /   (-1)  ------
       ---       (2n+1)!
       n=0

C(x) = cos(x)

        ∞
       ---           (2n)
C(x) = \       n    x
       /   (-1)   -----
       ---        (2n)!
       n=0

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