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This question already has an answer here:

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!

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

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    $\begingroup$ 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" $\endgroup$ – grand_chat Sep 2 '16 at 17:09
  • $\begingroup$ 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. $\endgroup$ – studiosus Sep 2 '16 at 17:52
  • $\begingroup$ @grand_chat, thanks for letting me know. $\endgroup$ – Gal Grünfeld Sep 2 '16 at 18:12
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$$\sum_{k=2}^{n}\frac{1}{k(k-1)} = \sum_{k=2}^{n}\frac{1}{k-1} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=\sum_{k=1}^{n-1}\frac{1}{k} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=1+\sum_{k=2}^{n-1}\frac{1}{k} - \sum_{k=2}^{n}\frac{1}{k}$$ $$=1-\frac{1}{n}+\sum_{k=2}^{n-1}\frac{1}{k} - \sum_{k=2}^{n-1}\frac{1}{k}$$

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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.

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    $\begingroup$ 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). $\qquad$ $\endgroup$ – Michael Hardy Sep 2 '16 at 17:08
  • $\begingroup$ Where is the induction? I told him the first step to make, the first step in every other answer posted. $\endgroup$ – Sean Lake Sep 2 '16 at 17:19
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$$\sum_{k=2}^n \frac{1}{k(k-1)} =\sum_{k=2}^n \underbrace{\frac{1}{k-1}}_{f(n-1)}-\sum_{k=2}^n \underbrace{\frac{1}{k}}_{f(n)}$$

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