Good idea, assuming that $\frac{x_n}{n}$ convergence I can show that your ideas work and your guess is correct.
Let's write $y_n = x_{n+1}-x_n$ so $x_n = \sum_{i=1}^n y_i$.
By assumption we have that $\sum_{i=n}^{n+k} y_i \rightarrow l$ as $n\rightarrow\infty$
we study the sub-sequence $x_{nk}$, we have that $$x_{nk} = \sum_{i=1}^{nk} y_i$$
$$\sum_{i=1}^{nk}y_i = \sum_{i=1}^k y_i + \sum_{i=k+1}^{2k}y_i+...+\sum_{i=(n-1)k+1}^{nk}y_i$$
Now denote the terms of this series by $z_1,z_2,...,z_n$ so $x_{nk}=\sum_{i=1}^n z_i$. Applying Cesaro Stolz lemma we have that $$\lim_{n\rightarrow\infty} \frac{x_{nk}}{nk} =\frac{1}{k}\lim_{n\rightarrow\infty} z_n$$
By assumption $z_n\rightarrow l$ as $n\rightarrow\infty$ so the limit is $l/k$.
We conclude that a subsequence of $\frac{x_n}{n}$ is converging to $l/k$. It is only left to show that the sequence convergence.
Edit: In fact, you can make similar argument for $x_{nk+1},...,x_{nk+k-1}$ and so you can 'cover' $x_n$ with sub-sequences converging to the same limit, this means that $x_n$ also converge to that limit.