Try the below method:change variables $b_n=C-a_n$,we get $b_{n+m}.Besides,it is easily seen that $\{ \frac{a_n}{n}\}$ and $\{\frac{b_n}{n}\}$ have the same convergent behavoir(both are convergent to some point,or both go to infinity,or both are divergent).The convergence of $\{\frac{b_n}{n}\}$ is a well known result.You can do it on your own or see as follows:let $b=\lim_{n\longrightarrow \infty}\inf_{k>n}\{ \frac{b_k}{k}\}$,$b$ exsits (may be infinity).Similiarly,we have $b'=\lim_{n\longrightarrow \infty}\sup_{k>n}\{ \frac{b_k}{k}\}$,and $b\leq b'$.Next we show that $b=b'$.By definition of $b$,$ \forall \epsilon >0,\exists N,s.t.|\frac{b_N}{N}-b|<\epsilon$.By definition of $b'$,there is a subsequnce $\{\frac{b_{n_r}}{n_r}\}$ which converges to $b'$.Note $\forall n_r,\exists k_r,q_r\in \mathbb{N}(0\leq q_r $.Beside $b_{n_r}$.Note $\{b_{q_r}\},\{q_r\}$ are bounded,thus taking limits with respect to $r$ in $\frac{b_{n_r}}{n_r}<\frac{k_r b_N+b_{q_r}}{k_r N+q_r}$,we get $b'\leq \frac{b_N}{N}$.Since $\epsilon $ is arbitary,we get $b'\leq b$.As a result,$b=b'$,which exactly means that the sequence $\{\frac{b_n}{n}\}$ converges.As it it unknown whether $b$ is finite or infinite,we get the two cases you have just pointed out.