# Evaluating sums question $\sum_{k=2}^n \frac{1}{k^2-1}$ [duplicate]

I need to use the identity $$\frac{1}{k^2-1}=\frac{1}{2}(\frac{1}{k-1}-\frac{1}{k+1})$$ to evaluate $$\sum_{k=2}^n \frac{1}{k^2-1}$$.

I am confused about how to begin this proof. Also, how am I supposed to use the identity here? Do I need to use induction?

• Write out several terms of the difference formatted version, and see that most "inner terms" cancel each other. – coffeemath Sep 15 '14 at 2:48
• This is a specific case of something more general, called a "telescoping series." – Akiva Weinberger Sep 15 '14 at 2:54
• The inner terms well cancel. Just the write the first few terms and you are going to figure out the form of cancellation easily. And yes you can use induction if you want, just find the $n$-th term and go for the inductive step to proof it for the $(n+1)$-th term. – Meshal Sep 15 '14 at 3:02
• – Martin Sleziak Dec 27 '19 at 6:20

Timbuc gave you the right way to approach the problem. If you take a deeper look to the results obtained for low values of $n$, you should easily find that the general term is just given by $$S_n=\sum_{k=2}^n\frac{1}{k^2-1}=\frac12\sum_{k=2}^n\left(\frac1{k-1}-\frac1{k+1}\right)=\frac{(n-1) (3 n+2)}{4 n (n+1)}$$
$$\frac12\sum_{k=2}^N\left(\frac1{k-1}-\frac1{k+1}\right)=$$
$$=\frac12\left(1-\color{red}{\frac13}+{\frac12}-\color{green}{\frac14}+\color{red}{\frac13}-\color{blue}{\frac15}+\color{green}{\frac14}-\frac16+\ldots+\color{purple}{\frac1{N-1}}-\frac1{N+1}\right)=\ldots$$