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ok, I will try that now... stuck on a problem

You could post it, if you wanted ¬_¬

ok, that might help...
\[\sum_{n=2}^{\infty} 2/n ^{2}-1\]

I understand how to do the partial fraction decomposition

I am getting stuck on the final terms when I sub in the series and cancel out

how do I arrive to the following answer....

\[s _{n}=1 + 1/2- (1/n-1) - (1/n)\]

so the last two terms in the last two parentheses stay?

or the last two iterations

not parentheses

I don't see how you are coming up with the last 3 terms with the n...??

See cancelling in image

As for the last terms, just sub in n-2, n-1 and n to the partial fraction
1/(n-1) - 1/(n+1)

I guess that is what I don't get... why am i subbing in n-2, n-1, and n? is that always?

ok, I think I understand.... thank you, Mr. Newton :)

No problem