Define $L := \lim_{n \to \infty} a_n$, $L' := \lim_{n \to \infty} (a_n + b_n)$.
Suppose that $\varepsilon > 0$. Then there exists $k \in \mathbb{N}$ such that for all $n \geq k,$ $|a_n - L| < \varepsilon/2$. There also exists a natural number $k'$ such that for all $n \geq k'$, $|a_n + b_n - L'| < \varepsilon/2$. Can you continue the argument from here?
Define $K := \text{max}(k, k')$, then for all $n \geq K$ we have that
\begin{align}|b_n - (L' - L)| &= |(a_n + b_n - L') - (a_n - L)| \\ &\leq |a_n + b_n - L'| + |a_n - L| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align} This shows that $b_n \to L' - L$ as $n \to \infty$.