Posted by Subhotosh Khan on September 13, 2002 at 12:39:18:
In Reply to: telescoping series posted by habaneroeater on September 13, 2002 at 11:33:03:
: Is 1/n(n+2) a telescoping series, if so, how do I convert this to 1/n-n/n+c form?
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I assume your correct problem was as follows:
1/[n*(n+2)]
In general, you will have to solve like this:
=A/n + B/(n+2)
=[A*(n+2) + B*n]/[n*(n+2)]
so
A*(n+2) + B*n = 1 ....from here you get
n(A+B) = 0
so A + B = 0 .....(1)
AND
2*A = 1 ...........(2)
Solve (1) and (2) to get the proper values.
However, this problem can be solved very easily by "observation" (which may not be possible all the time).....