>I'm not
>seeing how to get from 1/(nx(n-1)) to 1/(n-1) - 1/n. >
>Can you tell me if the actual steps are published anywhere
>that demonstrate how to get from point a to point b? I can
>even do it backwards, no problem, but don't see the
>intuitive method of getting from one to the other in the
>example indicated above.
Getting from b to a is as good as getting from a to b. The two are equal, after all.
However, there is a method of getting from a to b directly. I am sure you would rather prefer going from b to a. The pattern there is quite simple:
1/6 = 1/2 - 1/3
1/12 = 1/3 - 1/4, etc.
However, here's the method. It's called the method of "undetermined coefficients." It starts with an assertion that a fraction, like k/(m×n) can always be split into a sum of two fractions with denominators m and n. The numerators are to be found. Assume
k/(m×n) = x/m + y/n.
Then
k/(m×n) = (xn + ym)/(m×n),
so that
k = xn + ym;
You are of course looking for integer solutions. The only numbers k that can be represented in such a form are multiples of gcd(m,n). (You may want to search this site in particular or anywhere on the web for Euclid's algorithm.) In the case of the telescoping sequence, m = n-1, and, obviously, gcd(n-1,n) = 1. Thus you know that there are x and y such that
1 = xn + y(n-1).
But what might they be? It's almost immediate that x=1 and y=-1. In a more formal way you can try to use an "extension of Euclid's algorithm" (search this site.)