Trouble proving through induction after establishing a basecase What is the following sum?
$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + ... + \frac{1}{(n-1)n}$
Experiment, conjecture the value, and then prove it by induction.
I found the sum to be $\frac{n-1}{n}$ and established the base of n=2 being equal to $\frac 12$, but I don't know what to do next.

By induction assume that:$\frac 1 2 +\frac 1 6 +\frac {1}{12}+\dots \frac{1}{n(n-1)}=\frac{n-1}{n}$
For $n+1:$
$\underbrace{\frac 1 2 +\frac 1 6 +\frac {1}{12}+\dots \frac{1}{n(n-1)}}_{=\frac{n-1}{n}}+\frac{1}{n(n+1)}\stackrel{\color{red}?}{=}\frac{n}{n+1}$
$$\frac{1}{n(n+1)}+\frac{n-1}{n}=\frac{1+(n-1)(n-1)}{n(n+1)}=\frac{n^2}{n(n+1)}=\boxed{\frac{n}{n+1}}$$
As expected
