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

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!

## marked as duplicate by Arthur, user133281, 6005, Joey Zou, Parcly TaxelSep 3 '16 at 2:58

• Duplicate of math.stackexchange.com/q/1742825/215011 . But see the comments to that post: "...the telescoping method cannot be written out in all its excruciating fullness without the principle of induction" – grand_chat Sep 2 '16 at 17:09
• There is an easy algorithmic way for calculating hypergeometric sums, namely Gosper's algorithm. You can find it for instance in the book "A=B" by Zeilberger. – studiosus Sep 2 '16 at 17:52
• @grand_chat, thanks for letting me know. – Gal Grünfeld Sep 2 '16 at 18:12

$$\sum_{k=2}^{n}\frac{1}{k(k-1)} = \sum_{k=2}^{n}\frac{1}{k-1} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=\sum_{k=1}^{n-1}\frac{1}{k} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=1+\sum_{k=2}^{n-1}\frac{1}{k} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=1-\frac{1}{n}+\sum_{k=2}^{n-1}\frac{1}{k} - \sum_{k=2}^{n-1}\frac{1}{k}$$
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.
• I'm not really sure this can legitimately count as being "without induction" (although the induction is at most tacit when the argument is phrased like this). $\qquad$ – Michael Hardy Sep 2 '16 at 17:08
$$\sum_{k=2}^n \frac{1}{k(k-1)} =\sum_{k=2}^n \underbrace{\frac{1}{k-1}}_{f(n-1)}-\sum_{k=2}^n \underbrace{\frac{1}{k}}_{f(n)}$$