$\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c , c > 0$ . Then, is $\{x_n/n\}$ convergent? 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.
 A: 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 $x_n/n=\frac{1}{n}\sum_{k=1}^n a_k$.
Then
$$x_n/n-c=\left(\frac{1}{n}\sum_{k=1}^n a_k\right) - c = \frac{1}{n} \sum_{k=1}^n (a_k-c)\quad \color{red} { (2)  } $$
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.
Note:
In essence we just proved that a summable series is also Cesàro summable.
A: 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.
A: 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 $x_n$ ?
A: 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.
