Discrete Math Induction: $\sum^n_{i=1} \frac1{i(i+1)}$

Question

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?

Answer 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:

$$\begin{align}\sum_{i = 1}^{n + 1} \frac{1}{i(i+1)} &= \frac{1}{(n+1)(n+2)} + \sum_{i = 1}^{n} \frac{1}{i(i+1)}\\ &= \frac{1}{(n+1)(n+2)} + 1 - \frac{1}{n + 1}\\ &= 1 + \frac{1 - n - 2}{(n+1)(n+2)}\\ &= 1 - \frac{1}{n + 2}\end{align}$$