Is there a way to show the following equality without induction I wanted to show the following equality without using induction:
$$
\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}
$$
Any hint on how to do it?
 A: Hint: use $\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac 1k$.
A: Notice, $$\sum_{k=2}^{n}\frac{1}{k(k-1)}$$ $$=\sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)$$ $$=\left(\frac{1}{2-1}-\frac{1}{2}\right)+\left(\frac{1}{3-1}-\frac{1}{3}\right)+\left(\frac{1}{4-1}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{(n-1)-1}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)$$ $$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)$$ $$=\left(\frac{1}{1}-\frac{1}{n}\right)$$ $$=\frac{n-1}{n}$$
A: Hint: Partial fraction decomposition yields $$\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$$ which makes your series a special little beast called a telescoping series, try writing out the first few terms and see if you can spot a pattern that allows cancelling. 
In particular, you get $$\sum_{k=2}^n \frac{1}{k(k-1)} = \sum_{k=2}^n \left(\frac{1}{k-1} - \frac{1}{k}\right)$$ Which, when written out term by term looks something like $$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-2} - \frac{1}{n-1}\right) + \left(\frac{1}{n-1} - \frac{1}{n}\right)$$
Can you spot what cancels out?
A: Start by decomposing the fraction $\displaystyle\frac{1}{(k-1)(k)}$ into partial fractions. This can be done by setting the fraction equal to $\displaystyle\frac{A}{k-1}+\frac{B}{k}$, then substituting a two different values for $k$ into the equation $1=A(k)+B(k-1)$ (just multiplied both sides by the lowest common denominator), to get $A$ and $B$. Which will be $1$ and $-1$ respectively.
Now you have:
$$\sum\limits_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\dots+\frac{1}{(n-1)-1}-\frac{1}{(n-1)}+\frac{1}{(n)-1}-\frac{1}{n}$$
As you can see all the terms in the middle between $1$ and $(-1/n)$ will cancel out so that you're left with $1-\frac{1}{n}$, which you should know how to turn into the required form.
