Prove Without Induction: $\sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 1 - \frac{1}{n}$

Question

everybody.

I'm suppose to prove this without induction: Prove Without Induction: $\sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 1 - \frac{1}{n}$

I'm not sure how to do it. I tried a bit of algebraic manipulation, but I'm not sure how to do it.

It's suppose to be basic. I did get a hint of factorizing $\frac{1}{k(k-1)}$ but that didn't get me anywhere.

A hint or any directions would be much appreciated!

Answer 1

$$\sum_{k=2}^{n}\frac{1}{k(k-1)} = \sum_{k=2}^{n}\frac{1}{k-1} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=\sum_{k=1}^{n-1}\frac{1}{k} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=1+\sum_{k=2}^{n-1}\frac{1}{k} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=1-\frac{1}{n}+\sum_{k=2}^{n-1}\frac{1}{k} - \sum_{k=2}^{n-1}\frac{1}{k}$$

Answer 2

That's a telescoping series. Use partial fraction techniques to do the following split: $$\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k},$$ and proceed from there.

Answer 3

$$\sum_{k=2}^n \frac{1}{k(k-1)} =\sum_{k=2}^n \underbrace{\frac{1}{k-1}}_{f(n-1)}-\sum_{k=2}^n \underbrace{\frac{1}{k}}_{f(n)}$$