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Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.

$\textbf{My Attempt:}$ My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$ to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.

Any suggestions/hints would be greatly appreciated. Thanks!

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Let $a_n:=x_{n+1}-\frac{x_n}2$; then $a_n\to 0$. We have $$|x_{n+1}|=\left|a_n+\frac{x_n}2\right|\leqslant |a_n|+\frac{|x_n|}2.$$ Taking the $\limsup$, we get $$\limsup_{n\to \infty}|x_n|\leqslant \frac 12\limsup_{n\to \infty}|x_n|.$$ Take $M\geqslant 2$ and $n_0$ such that $|a_n|\leqslant 1$ if $n\geqslant n_0$, and assume that $|x_n|\leqslant M$. Then $|x_{n+1}|\leqslant M/2+1\leqslant M$. So the sequence $\{x_n\}$ is bounded and the $\limsup$ is finite.

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  • $\begingroup$ Why can we assume that $|x_n|\leq M$? $\endgroup$ – user81146 Oct 10 '13 at 19:03
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    $\begingroup$ @user81146 Take $M=2+|x_{n_0}|$. $\endgroup$ – Andrés E. Caicedo Oct 10 '13 at 19:06
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    $\begingroup$ Is the $M$ I suggested at least $2$? Is it at least $|x_{n_0}|$? What is there to expand? $\endgroup$ – Andrés E. Caicedo Oct 10 '13 at 19:41
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    $\begingroup$ There is a tricky argument you can use here: If $n=n_0$, then $n\ge n_0$. $\endgroup$ – Andrés E. Caicedo Oct 10 '13 at 20:09
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    $\begingroup$ Yes, exactly. That does it. (But you want $|x_{n_0}|$, $|x_{n_0+1}|$, etc -- don't forget the absolute values.) $\endgroup$ – Andrés E. Caicedo Oct 10 '13 at 20:35
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Hint: For any $\epsilon$ there is $N$ such that $\left|x_{n+1}-\dfrac{x_{n}}{2}\right| < \dfrac{\epsilon}{2}$ for any $n>N$. Then for any $m>N$ we have $$\left|x_{m}-\frac{x_{N}}{2^{m-N}}\right| = \left|\left(x_{m}-\frac{x_{m-1}}{2}\right)+\frac12\left(x_{m-1}-\frac{x_{m-2}}{2}\right)+\ldots+\frac1{2^{m-N-1}}\left(x_{N+1}-\frac{x_{N}}{2}\right)\right|< \epsilon.$$

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  • $\begingroup$ This is what I initially tried, but I can't seem to figure out how to get from $\left|x_m-\dfrac{x_n}{2^{m-n}}\right|<\epsilon$ to $\left|x_m-x_n\right|<\epsilon$. $\endgroup$ – user81146 Oct 10 '13 at 20:00
  • $\begingroup$ This inequality shows that $|x_m| < 2\epsilon$ for all sufficiently large $m$. Note that $\epsilon$ was arbitrary. $\endgroup$ – njguliyev Oct 10 '13 at 20:06

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