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Let $<x_n>$ be a bounded real number sequence. Show $\lim_n(x_{n+1} - x_n) = 0$ if $<x_n>$ satisfies $2x_n \le x_{n-1} + x_{n+1}$ for all $n \in \mathbb N$.

My try : Let $a_n = x_{n+1} - x_n$. The given condition is same as $x_n - x_{n-1} \le x_{n+1} - x_n$. Therefore, $<a_n>$ is monotonic, bounded, and convergent. If we can show $\sum_{n=0}^\infty a_n = 0$, then $lim_n(x_{n+1} - x_n) = 0$.

Now I'm stuck in how to show $\sum_{n=0}^\infty a_n = 0$. Is my approach right? If so, please give me an hint.

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  • $\begingroup$ This seems like a good start, but have you used that the sequences is bounded? $\endgroup$ Apr 12, 2017 at 11:36
  • $\begingroup$ @MichaelBurr Was not the boundedness of $x_n$ used to conclude that $a_n$ is bounded? $\endgroup$ Apr 12, 2017 at 11:48
  • $\begingroup$ I think you should make use of telescoping sums. $\endgroup$
    – mxian
    Apr 12, 2017 at 11:52
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    $\begingroup$ By the way, it is enough to show $\sum_n a_n < \infty$ to conclude $a_n \to 0$; one does not need to know the value of the sum. $\endgroup$ Apr 12, 2017 at 11:53
  • $\begingroup$ @LoveTooNap29 One approach is to use the boundedness again. $\endgroup$ Apr 12, 2017 at 11:53

1 Answer 1

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Hints:

  • You have a good start to observe that $x_n-x_{n-1}\leq x_{n+1}-x_n$.

  • Suppose that for some $m$, $x_m-x_{m-1}=\varepsilon>0$. Then, for all $n\geq m$, $x_n-x_{n-1}\geq\varepsilon>0$. This implies that $x_n\geq x_{m-1}+(n-m+1)\varepsilon$. This gives that $x_n$ is unbounded, a contradiction.

  • Therefore, $x_n$ is a decreasing sequence. Since bounded decreasing sequences are convergent, $x_n$ is a convergent sequence. But the the pairwise differences in convergent sequences go to zero.

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  • $\begingroup$ Thanks for detailed hint. You proved that $<x_n>$ is decreasing, bounded sequence. Therefore, $<x_n>$ is convergent, and it is cauchy sequence. As $<x_n>$ is a cauchy sequence, $lim_n(x_{n+1} - x_n) = 0$. Do I understand right? $\endgroup$
    – user432019
    Apr 12, 2017 at 12:50
  • $\begingroup$ Yes, that's right (although one doesn't need to go all the way to Cauchy, it works). $\endgroup$ Apr 12, 2017 at 12:55
  • $\begingroup$ Oh, I see. Thanks for your hint again. $\endgroup$
    – user432019
    Apr 12, 2017 at 13:05

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