<s>[INST] Prove $\frac12+\frac16+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$ for $n \in \mathbb{N}$ 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 $\frac{n+1}{(n+1)+1}$?
 [/INST] 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, $$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} = \frac{1}{1} - \frac{1}{n+1} = \frac{n}{n+1}.$$
</s>