Power Series

  ¥
  å   anxn = a1x + a2x2 + a3x3 + ... + anxn + ...  is an infinite series.
x=1

a1, a2, a3, … are the terms of the series.  an is the nth term.

If the sequence of partial sums has a limit S as n®¥, we say the series converges and we write
 ¥
 å    an  =  S
x=1
Otherwise, the series diverges.
 

Infinite Geometric Series

                               ¥
The geometric series å   anxn = ax + a2x2 + a3x3 + ... + anxn + ...
                      n=1

converges to the sum      a       if  | r | < 1 ,  and diverges if | r | > 1.
                                  1 - r
-1 < r < 1  is called the interval of convergence.
 

 ¥
 å   (2/7)k  is a geometric series with r = 2/7, and |2/7| < 1, so the sum is
k=1
                        2/7       =    2/7   =   2
                    1 - (2/7)         5/7        5

  The series converges to 2/5.

Examples:

1)  -2 + 2 – 2 + 2 – 2 + …
     The sequence of partial sums is  -2, 0, -2, 0, -2, 0, ....
     This sequence does not approach a particular number, so there is no sum.  It diverges.

2)  .2 + .02 + .002 + .0002 + …
     This is an infinite geometric series with r = .1,   |.1| < 1, so the sum is

         .2     =  .2   =   2
       1 - .1      .9        9

      The series converges to 2/9.

3)  1 – ½ + ¼ - 1/8 + … + (-1/2)n-1 + ...
     This is an infinite geometric series with r = -1/2,  |-1/2| < 1, so the sum is

                       1               =     1       =     2
                  1 - (-1/2)              3/2            3

   The series converges to 2/3.

4)  1 + 2 + 3 + 4 + … + n + …
     The sequence of partial sums for this series is 1, 3, 6, 10, ..., n(n + 1), ...
                                                                                                   2
     This sequence just continues to grow, so the sum does not exist.  It diverges.
 

Functions can be represented by series.

If | x | < 1, then 1 + x + x2 + … + xn + … =     1      since this is a geometric series with r = x.
                                                                      1 – x
Find a power series that represents            3
                                                             1 – x3
   In this geometric sum, a = 3 and r = x3.

   So,      3      =   3 + 3x3 + 3x6 + 3x9 + ... + 3x3n + ..., n > 0   and   | x3| < 1
          1 – x3
 

Power Series

¥
  å   cn(x - a)n = c0 + c1(x-a) + c2(x-a)2 + c3(x-a)3 + ... + cn(x-a)n + ...  is a power series
x=0
         centered around x = a.

        The domain is the interval over convergence.
        1.  The series converges only at a.
        2.  There exists a real number R > 0 such that the series converges (absolutely) for
             | x - a | < R,    and diverges for | x - a | > R.
        3.  The series converges for all x.
 

Differentiation

If f is represented by a series and that series converges for | x – a| < R, then f ‘ represented by the derivative of that series (differentiating term by term) converges for |x – a| < R.

If the series for f converges for all x, then so does the series for f ‘.
 

Integration

If f is represented by a series and that series converges for |x – a| < R, then F represented by the integral of that series (integrating term by term) converges for |x – a| < R.

If the series for f converges for all x, then so does the series for F.

Use differentiation to find a series for f(x) = 2/(1 – x)3.

   f(x) =     1      =  1 + x + x2 + x3 + x4 + ... + xn + ... ,  for n > 0 and |x| < 1
              1 - x

   f '(x) =      -1         = 1 + 2x + 3x2 + 4x3 + ... + nxn-1 + ...., for n > 1 and |x| < 1
                (1 - x)2

  f ''(x) =      2          =    2 + 6x + 12x2 + ... + n(n - 1)xn-2 + ..., for n > 2 and |x| < 1
               (1 - x)3
 
 

Now, obtain a series for ln (1 + x) by integrating the series for      1     .
                                                                                                1 + x

    1        =   1 - x + x2 - x3 + x4 - .... (-1)n-1xn-1 + ... ,  for n > 1 and |-x| < 1
 1 + x

                  ¥
ln (1 + x) = ò  1/(1 + x) dx =  x - x2 + x3 - x4 + x5 - ... + (-1)n-1xn + ..., for n > 1 and |-x| < 1
                 1