<s>[INST] prove formula by induction I need to prove by induction that $\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$.
I've seen some similar questions about the convergence of the infinite series, however none about the finite case, I've tried in several ways to factorize or add things in order the find the case m+1 but without success.
 [/INST] Note:  If
$\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$
Then $\sum_{n=2}^{m+ 1} \frac{1}{n^2 - 1} = (\sum_{n=2}^{m}  \frac{1}{n^2 - 1}) + \frac {1}{(m+1)^2 - 1} =$
$(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{(m^2 + 2m +1)-1}=$
$(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{m(m + 2)}$
Can you finish?
Hint:  $\frac 1k - \frac 1{k+a} = \frac {k+a}{k(k+a)} - \frac {k}{k(k+a)} = \frac {(k+a)-k}{k(k+a)} = \frac {a}{k(k+a)}$.
Hint 2: Equality goes both ways.
Hint 3:You did put "telescoping-series" as a tag.....
Hint 4:  You know that you've got to end up with $\frac 12(1+\frac 12 -\frac 1{m+1} - \frac 1{m+2})$....  Can you show $(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{m(m + 2)} = \frac 12(1+\frac 12 -\frac 1{m+1} - \frac 1{m+2})$.  If you can you are done.  If you can't you're dinked.
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