Prove for the base case (usually $n=0$ or $n=1$) and show that if it is true for $k$, then it must be true for $k+1$. For example, let's use proof by induction to prove $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$ For the base case where $n = 1$ we can show $\displaylines{\frac{1}{i(i+1)} = \frac{1}{1(1+1)} = \frac{1}{1+1} = \frac{n}{n+1} \\ \frac12 = \frac12}$ We will assume that it is true for $k$ (e.g., where $n = k$) $\sum_{i=1}^{k}\frac{1}{i(i+1)} = \frac{k}{k+1}$ We then need to show that it is true for $k+1$. In other words, we need to show $\sum_{i=1}^{k+1}\frac{1}{i(i+1)} = \frac{k+1}{k+2}$ We can expand the sum as follows by substituting $k+1$ for $i$ for the last term in the sum. $\sum_{i=1}^{k+1}\frac{1}{i(i+1)} = \sum_{i=1}^{k}\frac{1}{i(i+1)} + \frac{1}{(k+1)(k+2)}$ By our initial assumption that this is true for $k$, we can substitute the summation here with $\frac{k}{k+1}$ to get $\sum_{i=1}^{k+1}\frac{1}{i(i+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$ Factoring the denominator we get $\sum_{i=1}^{k+1}\frac{1}{i(i+1)} = \frac{(k+2)k}{(k+2)(k+1)} + \frac{1}{(k+1)(k+2)}$ Notice that $(k+2)k + 1 = k^2 + 2k + 1 = (k+1)^2$ ) to simplify the above to get $\sum_{i=1}^{k+1}\frac{1}{i(i+1)} = \frac{k+1}{k+2}$