Discrete Math Induction: $\sum^n_{i=1} \frac1{i(i+1)}$ For $\sum^n_{i=1} \frac1{i(i+1)}$ Find a formula and proofs that it holds for all n ≥ 1.
How would I find the formula for this one that can hold for all n ≥ 1?  

Notice that:
$$\frac{1}{i} - \frac{1}{i + 1} = \frac{1}{i(i+1)}$$
Afterwards, observe the cancellations that occur in:
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \dots - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1}\\
= 1 - \frac{1}{n + 1}$$
Then, we have to prove this by induction (although I find it unnecessary): Here's the inductive step:
$$\begin{align}\sum_{i = 1}^{n + 1} \frac{1}{i(i+1)} &= \frac{1}{(n+1)(n+2)} + \sum_{i = 1}^{n} \frac{1}{i(i+1)}\\
&= \frac{1}{(n+1)(n+2)} + 1 - \frac{1}{n + 1}\\
&= 1 + \frac{1 - n - 2}{(n+1)(n+2)}\\
&= 1 - \frac{1}{n + 2}\end{align}$$
