let $v_n = \frac{a_n}{m+n}$, and $u_n = \frac{a_n}{n}$. Now observe that $\frac{a_n}{m+n} \le \frac{a_n}{2n}$, so by comparison test, we can say that as each term of $v_n$ is smaller or equal to that of $\frac{u_n}{2}$, and as $u_n$ converges, so $v_n$ also converges as $m \to \infty$.
I think this is it.
:)