I am using Induction:
Base Case $n=1$ holds ; $\frac12$= $\frac{1}{(1)+1}$
Assume $\frac{n}{n+1}$ is true from some $n \in \mathbb{N}$.
Then $\frac12+\frac16+...+\frac{1}{n(n+1)}+ \frac{1}{(n+1)((n+1)+1)}=\frac{n}{n+1}+ \frac{1}{(n+1)((n+1)+1)}$. By the Inductive Hypothesis
from here do I simplfy the RHS showing that it equals $\frac{n+1}{(n+1)+1}$?
$$\frac n{n+1}+\frac1{(n+1)(n+2)}=\frac{(n^2+2n)+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}$$
For completeness sake, we evaluate the sum without induction: Note that $$ \frac{1}{k(k+1)} = \frac{(k+1) - k}{k(k+1)} = \frac{k+1}{k(k+1)} - \frac{k}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.$$ Therefore, $$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} = \frac{1}{1} - \frac{1}{n+1} = \frac{n}{n+1}.$$
Well, I will tell you a method without having to use induction. We have, $$\frac {1}{2} = \frac {2-1}{2\times 1} = \frac {1}{1}-\frac {1}{2} $$ $$\frac {1}{6} = \frac {3-2}{3\times 2} = \frac {1}{2}-\frac {1}{3} $$ $$\vdots $$ $$\frac {1}{n (n+1)} = \frac {n+1-n}{n (n+1)} = \frac {1}{n}-\frac {1}{n+1} $$
Adding, we get, $$\frac {1}{2}+\frac {1}{6}+\cdots + \frac {1}{n(n+1)} = (\frac {1}{1}-\frac {1}{2})+(\frac {1}{2}-\frac {1}{3})+\cdots + ( \frac{1}{n}-\frac {1}{n+1}) = \frac {1}{1}-\frac {1}{n+1} = \frac {n}{n+1} $$
This is also called a telescoping sum. Hope it helps.