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Let ${x_n}$ be a real sequence such that $\lim_{n\to\infty}(x_{n+1}-x_n)=c$. Then, talk about convergence of the sequence ${\frac{x_n}{n}}$

My try: I did not understand how to proceed. I thought of integers and even integers. The former's seq $\frac{x_n}{n}$ converges to 1 and latter to 2.

So i think, maybe, the sequence should converge to $c$. But i have no clue how to prove it. Any hwlp would be appreciated.

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  • $\begingroup$ You know that lim x_n+1=lim x_n=x. (Subsequences go to the same limit). This c=x-x=0 assuming that $x_n$ is convergent (which it has to then c won't be a number then). From here you can get the limit for $\frac{x_n}{n}$ $\endgroup$ – user60887 Jun 8 '14 at 5:37
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Hint: If $U > x_{n+1} - x_n > L$ for all $n \ge N$, what can you say about $x_n - x_N$?

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  • $\begingroup$ $=0$ for $n=N$, else $>L$ $\endgroup$ – user141561 Jun 8 '14 at 5:53
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Solution 1: Use Cesàro mean.

Solution 2: Roughly speaking, $(x_n)$ looks like an arithmetic sequence, $x_{n+1} \approx x_n+c$. Therefore, $x_n \approx nc$ hence $\frac{x_n}{n} \approx c$. Finally, you just have to explicit the approximations to conclude.

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