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### Differences Method

```
Date: 07/06/2001 at 05:36:29
From: Mark Atkinson
Subject: The method of differences

How does the method of differences work?
```

```
Date: 07/06/2001 at 05:51:55
From: Doctor Anthony
Subject: Re: The method of differences

To find the sum of n terms and the sum to infinity of the series

1         1         1
----  +  ------  + ------ + ........
1 x 2     2 x 3     3 x 4

You use the method of differences.

u(n) = 1/(n(n+1))      =  1/n     -  1/(n+1)
u(n-1)= 1/((n-1)n)     =  1/(n-1) -  1/n
u(n-2)= 1/((n-2)(n-1)) =  1/(n-2) -  1/(n-1)
............................................
............................................
u(3) = 1/((3)(4))      =  1/3     -  1/4
u(2) = 1/((2)(3))      =  1/2     -  1/3
u(1) = 1/((1)(2))      =  1/1     -  1/2
SUM[u(r)]                = 1  - 1/(n+1)
r = 1 to n

When you add all the equations you get cancellation between rows, and
everything goes out except the first term of u(1) and the last term of
u(n).

So sum of the series is  1 - 1/(n+1)

as n -> infinity the term 1/(n+1) -> 0, and so the sum of the series
to infinity is 1.

SUM[u(r)]        = 1
(r = 1 to infinity)

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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