prove formula by induction

Question

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.

Answer 1

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.

Answer 2

Here is an explicit evaluation which may be useful. This is focused on the $m+1$ portion of the inductive proof:

$P\left(m+1\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m+1}-\frac{1}{m+2}\right)$

$P\left(m+1\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m+1}\right)-\frac{1}{2}\frac{1}{m+2}$

In order for the inductive step to be valid, the above expression needs to be equal to:

$ P\left(m\right)+\frac{1}{\left(m+1\right)^{2}-1}=P\left(m\right)+\frac{1}{m^{2}+2m} $

where $P\left(m\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1}\right)$ is assumed true.

We thus require that:

$ \frac{1}{2}\frac{1}{m+2}=\frac{1}{2m+4}=\frac{1}{2m}-\frac{1}{m^{2}+2m} $

where the first term on the right completes the expression for $P\left(m\right)$ and the second term is the new increment of $\frac{1}{n^{2}-1}$ for $m+1$.

Cross-multiplying the right hand side of the above yields:

$ \frac{1}{2m+4}=\frac{m^{2}}{2m^{3}+4m^{2}} \checkmark $

I hope this helps.