Domain = Interval of Convergence

                                                                                               ¥
Only one of three things can happen for any series of the form  S      cn(x - a)n
                                                                                            n=0
The number R is called the radius of convergence.

1.  The series converges only at a. (R = 0)
2.  There exists a real number R > 0 such that the series converges (absolutely) for |x - a| < R,
     and diverges for |x - a| > R.  The series may or may not converge at the endpoints.
3.  The series converges for all x.  (R = ¥)

The set of all values of x for which the power series converges is called the interval of convergence of the power series.
 

Nth term test

San diverges if     lim     an fails to exist or is not 0.
                          n®¥

                                                                        ¥
Example:  Find the radius of convergence for   S    n!xn
                                                                                                n=0

   If x > 1 or x < -1, the terms grow without bound as n®¥, and the series diverges by the nth
   term test.

   If |x| < 1, then   lim     n!xn   =     lim    n! / (1/x)n
                         n®¥                  n®¥
   Factorial functions grow faster than power functions, so this limit does not exist.

   If x = 0, the series does converge.  The radius of convergence is 0 and the interval of
   convergence is x = 0.
 

Direct Comparison Test

Let San and Scn be a series with no negative terms.
If Scn converges, then San converges if an < cn for all n > N for some integer n.
If Scn diverges, then San diverges if an> cn for all n > N for some integer n.

Example:

      ¥
     S    1 / (1 + 2n)
    n=0

   Compare to  ¥
                      S      1 / 2n
                    n=0
   This is a geometric series with |r| = 1/2 < 1, so the series converges.
   Since  ¥                         ¥
           S     1 / (1 + 2n)  <  S      1 / 2n  term for term, the first series also converges.
           n=0                            n=0
 
 

Absolute Convergence

If the series å |an| converges, then the series å an also converges.
1.  å anis absolutely convergent if å |an| converges.
2.  å an is conditionally convergent if å an converges but å |an|diverges.

Proof:
Since 0 < an + |an| < 2|an| for all n, the series

             ¥
           S     an + |an|
           n=1

converges by comparison to the convergent series 2|an|.  Also, since an = (an + |an|) - |an|, we can write
            ¥            ¥                     ¥
           S     an  =  S   ( an + |an|)   -   S    | an|
           n=1              n=1                      n=1

where both series on the right converge.   Hence å an converges.
 

Example:

    ¥
  å   (-1)n / 3n
n=1

This series converges absolutely because
   ¥
  å   1/ 3n   converges since this is a geometric series with r = 1/3 .
n=1
 
 

Ratio Test:

Let å an be a series with nonzero terms.
1. å an  converges if  lim      |an+1 |   <   1
                                 n®¥   |  an   |

2. å an  diverges if   lim      |an+1 |   >   1
                              n®¥   |  an   |

3. Inconclusive if   lim      |an+1 |   =   1
                          n®¥   |  an   |

This is a good test to use for series that converge rapidly (involving factorials or exponents).

Example:

    ¥
  å   2n / n!
n=1

  lim      |    2n+1     n!|   =   lim        2       =   0   <   1
n®¥   |(n + 1)!   2n |       n®¥  n + 1

Thus, by the Ratio Test, the series converges.
 

Telescoping Series:

These are series whose sequence of partial sums show terms being subtracted out.

Example:
    ¥
  å   1/ [n(n + 1)]
n=1

Using partial fractions,  1 / [n(n + 1)] =  1/n  -  1/(n + 1)

    ¥
  å   [ 1/ n   +   1 / (n + 1) ]   =  1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - ....   =   1
n=1

The series converges to 1.