Simple expression for $\sum_{k=1}^{n-1}\:\frac{1}{k\left(k+1\right)}$

Question

I know that

$\:\:\frac{1}{k\left(k+1\right)}\:\:\:\:=\:\frac{1}{k}\:-\:\frac{1}{k+1}\:$

And that

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

But I'm not completely sure how to turn $\sum_{k=1}^{n-1}\:\frac{1}{k\left(k+1\right)}$ into a simple expression using this information.

Answer 1

The second information is unrelevant here.

Note that $$\sum_{k=1}^{n-1}\:\frac{1}{k\left(k+1\right)} = \sum_{k=1}^{n-1} \frac{1}{k}\:- \sum_{k=1}^{n-1} \:\frac{1}{k+1}.$$ Now do a reindexation of the second sum and you will get $$\sum_{k=1}^{n-1} \frac{1}{k}\:- \sum_{k=1}^{n-1} \:\frac{1}{k+1} = \sum_{k=1}^{n-1} \frac{1}{k}\:- \sum_{k=2}^{n} \:\frac{1}{k} = 1-\frac{1}{n}.$$

Answer 2

It telescopes, write down the first few terms using $\frac1k-\frac1{k+1}$ and you'll see most terms cancel.