real analysis - Limit of $b_n$ when $b_n=frac{a_n}{n}$ and $limlimits_{ntoinfty}(a_{n+1}-a_n)=l$

I have the following problem.

I have to find the limit of $b_n = \frac{a_n}{n}$, where $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$

My approach:

I express $a_n$ in terms of $b_n$, i.e.
$a_n=nb_n$ and $a_{n+1}=(n+1)b_n$

We look at the difference: $a_{n+1}-a_n=(n+1)b_{n+1}-nb_n$

Assuming that $b_n$ converges to a real number m, we see that:

$l=(n+1)m-nm$, from where I conclude that $m=l$.

What I'm left with is proving that $b_n$ is convergent which I'm not sure how to do.

Thanks in advance!

Answer

As @ParamanandSingh suggested, from Stolz–Cesàro theorem
$$a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n} \rightarrow l, n \rightarrow \infty$$
where $\{n\}_{n \in \mathbb{N}}$ is monotone and divergent, then
$$\frac{a_n}{n} \rightarrow l, n \rightarrow \infty$$