calculus - Find $limlimits_{n to infty} frac{x_n}{n}$ when $limlimits_{n to infty} x_{n+k}-x_{n}$ exists

Let $(x_n)_{n \geq 1}$ be a sequence with real numbers and $k$ a fixed natural number such that

$$\lim_{n \to \infty}(x_{n+k}-x_n)=l$$

Find

$$\lim_{n \to \infty} \frac{x_n}{n}$$

I have a strong guess that the limit is $\frac{l}{k}$ and I tried to prove it using the sequence $y_n=x_{n+1}-x_n$. We know that $\lim_{n \to \infty}(y_n+y_{n+1}+\dots+y_{n+k-1})=l$ and if we found $\lim_{n \to \infty}y_n$ we would have from the Cesaro Stolz lemma that

$$\lim_{n \to \infty}\frac{x_n}{n}=\lim_{n \to \infty}y_n$$

Answer

For fixed $m \in \{ 1, \ldots, k \}$ the sequence $(y_n)$
defined by $y_n = x_{m+kn}$ satisfies

$$y_{n+1} - y_n = x_{(m+kn) + k} - x_{m+kn} \to l \, ,$$
so that Cesaro Stolz can be applied to $(y_n)$. It follows that $\frac{y_n}{n} \to l$ and
$$\frac{x_{m+kn}}{m+kn} = \frac{y_n}{n} \cdot \frac{n}{m+kn} \ \to \frac{l}{k} \text{ for } n \to \infty \, .$$
This holds for each $m \in \{ 1, \ldots, k \}$, and therefore
$$\lim_{n \to \infty} \frac{x_n}{n} = \frac lk \, .$$