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

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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 $$ \lim_{n \to \infty} \frac{x_n}{n} = \frac lk \, . $$

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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.

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Assume $y_n = x_{n+1}-x_n\to \ell$, then by cesaro summation we have and telescopic sum we have,

$$\ell= \lim_{n\to\infty}y_n=\lim_{n\to\infty}\frac{1}{n}\sum_{k}^{n}y_k = \lim_{n\to\infty}\frac{ 1}{n}\sum_{k}^{n}x_{k+1}-x_{k} =\lim_{n\to\infty}\frac{x_{n+1}-x_{1}}{n}= \lim_{n\to\infty}\frac{x_{n+1}}{n}$$Hence $$\ell=\lim_{n\to\infty}\frac{x_{n+1}}{n}= \lim_{n\to\infty}\frac{x_{n}}{n}$$

Generally, $k\neq 1$ $y_n = x_{n+k}-x_n\to \ell$ and we assume, existence of $\lim_{n\to\infty}\frac{ x_{n+1}}{n} $, $$\ell= \lim_{n\to\infty}y_n=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}y_i = \lim_{n\to\infty}\frac{1}{n} \sum_{i=1}^{n}x_{i+k}-x_{i} = \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{k}\sum_{i=1}^{n}x_{i+j+1}-x_{i+j} \\=\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{k} (x_{n+j+1}-x_{j})= \lim_{n\to\infty}\frac{k x_{n+1}}{n} $$hence
$$\frac{\ell}k=\lim_{n\to\infty}\frac{x_{n}}{n}$$

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    $\begingroup$ How do you know that $y_n \to l$? We only know $x_{n+k}-x_n \to l$, which is different. $\endgroup$
    – AndrewC
    Commented Feb 9, 2018 at 20:35
  • $\begingroup$ @AndrewC Idid it for k=1 I am editing the general case $\endgroup$
    – Guy Fsone
    Commented Feb 9, 2018 at 20:36
  • $\begingroup$ I may be wrong, but it seems to me that you assume the existence of $\lim_{n\to \infty}\frac{x_{n+1}}{n}$ in $\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{k} (x_{n+j+1}-x_{j})= \lim_{n\to\infty}\frac{kx_{n+1}}{n}$ $\endgroup$
    – Martin R
    Commented Feb 9, 2018 at 21:11
  • $\begingroup$ @MartinR you totally right $\endgroup$
    – Guy Fsone
    Commented Feb 9, 2018 at 21:13
  • $\begingroup$ May the down voter explain himself $\endgroup$
    – Guy Fsone
    Commented Feb 10, 2018 at 9:21

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