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.
On
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
You established that it worked for $n=2$ (since $\frac1{1\cdot2}=\frac12=1-\frac1n$. This was the base case). Now assume that the statement is true for some $n=k$, so we have $\sum_{i=2}^k\frac1{i(i-1)}=1-\frac1k$ (this is the inductive hypothesis). So we have \begin{align} \sum_{i=2}^{k+1}\frac1{i(i-1)}&=\frac1{(k+1)((k+1)-1)}+\sum_{i=2}^k\frac1{i(i-1)}\\ &\stackrel{IH}{=}\frac{1}{k(k+1)}+1-\frac1k\\ &=\frac 1{k}-\frac1{k+1}+1-\frac 1k\\ &=1-\frac1{k+1}\\ \end{align} In the first step, we take the term with $i=k+1$ out of the summation. Then the $\stackrel{IH}{=}$ sign indicates that, in that very step, we used our induction hypothesis. Now we proved that, if the statement is true for $k$, then also for $k+1$; and since it was true for $n=2$, it must be true for $3$, and this also for $4$, and for $5$, etc., and therefore it must be true for all natural numbers $n\geq 2$ (this is the principle of mathematical induction).
Alternatively, we can do this without induction by doing the following: \begin{align} \sum_{i=2}^n\frac{1}{i(i-1)}&=\sum_{i=2}^n\left(\frac 1{i-1}-\frac1i\right)\\ &=\left(\sum_{i=2}^n\frac1{i-1}\right)-\left(\sum_{i=2}^n\frac 1i\right)\\ &=\left(\sum_{i=1}^{n-1}\frac1i\right)-\left(\sum_{i=2}^n\frac1i\right)\\ &=\frac11-\frac1n+\sum_{i=2}^n\frac1i-\sum_{i=2}^n\frac1i\\ &=1-\frac1n \end{align}