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Let $(x_n)_{n\geq 1} \subset \left[-\frac{1}{2}, \frac{1}{2} \right]$ such that $|x_m+x_n|\geq \frac{m}{n}, \forall m,n \in \mathbb{N}_{\geq1 }$ with $m<n$. Prove that $(x_n)$ converges.

We have $1=\frac{1}{2}+\frac{1}{2} \geq |x_{n}|+|x_{n-1}| \geq |x_n+x_{n-1}|\geq \frac{n-1}{n} \to 1$, so $$\lim_{n\to \infty} |x_n+x_{n-1}|=1$$ So we should prove that $x_n \to \frac{1}{2} \text{ or } -\frac{1}{2}$, but I didn't succeed in proving it.

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    $\begingroup$ Can you see that what you have implies that the sequence cannot have an accumulation point except $\pm \frac{1}{2}$? If the sequence has only one accumulation point, it is convergent. So assume that both, $\frac{1}{2}$ and $-\frac{1}{2}$ were accumulation points … $\endgroup$ – Daniel Fischer Oct 31 '17 at 17:23
  • $\begingroup$ Yes, thank you! $\endgroup$ – Shroud Oct 31 '17 at 17:40
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  1. For all $n \ge 3$, the condition $|x_n + x_{n-1}| \ge \frac{n-1}{n} > \frac{1}{2}$ implies that $x_n$ and $x_{n-1}$ must have the same sign. So all (except maybe one) terms in the sequence have the same sign.
  2. Without loss of generality assume all terms of the sequence are positive. Then we claim the limit is $\frac{1}{2}$. (In the other case where the $x_n$ are negative, the limit is $-\frac{1}{2}$.) Why?

Suppose for sake of contradiction that $x_n < \frac{1}{2} - \epsilon$ for infinitely many $n$. Then for such $n$, we have $|x_n + x_{n-1}| < 1-\epsilon$, which contradicts $|x_n + x_{n-1}| \to 1$. (The other case when the $x_n$ are negative can be handled similarly.)

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With $|x_{n+1}|\geq|x_{n+1}+x_{n}|-|x_{n}|$, we have $\liminf_{n}|x_{n+1}|\geq 1-\limsup_{n}|x_{n}|\geq 1-\dfrac{1}{2}=\dfrac{1}{2}$, so $\dfrac{1}{2}\leq\liminf_{n}|x_{n}|\leq\limsup_{n}|x_{n}|\leq\dfrac{1}{2}$. Now As @angryawin has noted, then $x_{n}$ have all the same positive sign, then $\lim_{n}x_{n}=\dfrac{1}{2}$.

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