Sum of Convergent Series

Date: 09/25/1999 at 23:49:18
From: Abby
Subject: Sum of a convergent series

Could someone tell me how to find the sum of a convergent series such 
as these:

     infinity
        Sum  1/[(k+1)(k+3)]
       k = 0

     infinity
        Sum  [(25/10^k) - (6/100^k)]
       k = 0

Thanks in advance.

Date: 09/26/1999 at 08:14:24
From: Doctor Anthony
Subject: Re: Sum of a convergent series

>     infinity
>        Sum  1/[(k+1)(k+3)]
>       k = 0


         1         A     B
     ---------- = --- + ---
     (k+1)(k+3)   k+1   k+3

     1 = A(k+3) + B(k+1)

When k = -1:   1 = 2A    and so   A = 1/2
     k = -3:   1 = -2B   and so   B = -1/2

                1          1        1
Therefore   ---------- = ------ - ------
            (k+1)(k+3)   2(k+1)   2(k+3)

Our series is therefore the sum of the following terms:

     1/[2(1)] - 1/[2(3)] +
     1/[2(2)] - 1/[2(4)] +
     1/[2(3)] - 1/[2(5)] +
     1/[2(4)] - 1/[2(6)] +
        :          :

     and so on to infinity.

We can see that terms will start to cancel between lines except the 
first terms on the first and second lines. So the sum to infinity is:

     1/[2(1)] + 1/[2(2)]   =  1/2 + 1/4  =  3/4


>     infinity
>        Sum  [(25/10^k) - (6/100^k)]
>       k = 0

We could split this into two series

     25[1 + 1/10 + 1/10^2 +  ...]   - 6[1 + 1/100 + 1/100^2 + ...]

These are both geometric series, and using the formula a/(1-r) as sum 
to infinity

    25/[1-(1/10)] - 6/[1-(1/100)] = 250/9 - 600/99

                                  = 2150/99

                                  = 21.717171...

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/