**Proof**: Take $\epsilon>0$. Since $a_{n} \to a$ and $b_{n} \to b$, $\exists N_{1}, N_{2} \in \mathbb{N}$ such that for $n \geq \max(N_{1},n_{2})$ $|a_{n}-a|< \frac{\epsilon}{2} \text{ and } |b_{n}-b|< \frac{\epsilon}{2}$so using [[Absolute Value Satisfies Triangle Inequality]], $|(a_{n}+b_{n})-(a+b)|\leq |a_{n}-a|+|b_{n}-b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$so $a_{n}+b_{n} \to a+b$ as claimed.