If $\{x_n\}$ is a sequence which satisfies $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$ where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of $\{x_n/n\}$.

Attempt: $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$ where $c >0$

=> $x_n$ is unbounded and divergent.

However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.

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    $\begingroup$ Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=\sum_{k=1}^n a_k$. So, the assumption is that $a_n\to c$ and we are looking for the limit of the sequence of averages of $a_k$'s. $\endgroup$ – Berci Feb 9 '14 at 18:27
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    $\begingroup$ This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one. $\endgroup$ – Martin Sleziak May 31 '14 at 12:01

We claim that $x_n/n\rightarrow c$.

As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=\sum_{k=1}^n a_k$ $\color{red} { (1) }$. The assumption becomes $a_n\rightarrow c$ and $x_n/n=\frac{1}{n}\sum_{k=1}^n a_k$.


$$x_n/n-c=\left(\frac{1}{n}\sum_{k=1}^n a_k\right) - c = \frac{1}{n} \sum_{k=1}^n (a_k-c)\quad \color{red} { (2) } $$

Let $\epsilon>0$ and choose $N$ large enough such that $|a_n-c|<\epsilon$ for all $n\ge N$. Then

$$\left|\frac{x_n}{n}-c\right|\le \frac{1}{n}\sum_{k=1}^N |a_k-c| + \frac{n-N}{n}\epsilon$$

for $n\ge N$. Now let $n\rightarrow\infty$ while keeping $N,\epsilon$ fixed. Then we obtain

$$\limsup_{n\rightarrow\infty}\left|\frac{x_n}{n}-c\right|\le \epsilon$$

Since $\epsilon$ was arbitrary, the conclusion follows.


In essence we just proved that a summable series is also Cesàro summable.

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That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$

$$\lim_{n \to \infty} x_{n+1}-x_n = \lim_{n \to \infty} \frac{x_n}{n}$$

when $\lim_{n \to \infty} x_{n+1}-x_n$ exists.

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Apologies for editing Your Ad Here, but can someone please check (1) and (2)?

(1) = $\sum_{1 \le k \le n} x_k - x_{k - 1}$ which telecopes to $-x_0 + x_n$, not just $x_n$ ?

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  • $\begingroup$ That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,\dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer. $\endgroup$ – J.R. May 26 '14 at 20:05
  • $\begingroup$ @YourAdHere thanks $\endgroup$ – Group Theory May 29 '14 at 20:31

We let $a_n:=x_{n+1}-x_n$ and use this theorem to conclude that $$\frac{a_1+a_2+...+a_n}{n}\to c \implies \frac{(x_2-x_1)+(x_3-x_2)+....+(x_{n+1}-x_n)}{n}\to c $$ $$ \implies \frac{x_{n+1}-x_1}{n}\to c \implies \frac{x_n}{n}\to c$$ since $x_1$ is finite.

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