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Let $(x_n)$ be a sequence in $\mathbb{R}$. Show that if $(2x_{n+1}-x_n)$ converges to $x$, then $(x_n)$ converges to $x$.

We don't know if the sequence $(x_n)$ is a convergent sequence or a cauchy sequence. If we are able to prove any one, then the problem is simple. Please provide any clue on how to go about.

Shreedhar Bhat
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2 Answers2

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Hint:

\begin{align} |x_m-x_n| =&\ \frac{1}{2}|2x_m-x_{m-1}+x_{m-1}-x_{n-1}+x_{n-1}-2x_n|\\ \leq&\ \frac{1}{2} |(2x_m-x_{m-1})-(2x_n-x_{n-1})|+\frac{1}{2}|x_{m-1}-x_{n-1}|. \end{align}

Probably a very bad hint!

Jacky Chong
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  • I would need another hint :p Because if $a_{n, m}$, $b_{n ,m}$ are nonnegative real sequences with $a_{n, m} \leq \frac12 b_{n, m} + \frac12 a_{n-1,m-1}$ and $b_{n, m} \to 0$, it need not be the case that $a_{n, m} \to 0$. Take for example $b = 0$ and $a_{n, m} = 2^{n-2m}$ for $n \geq m$, zero otherwise. – Bart Michels Aug 05 '18 at 07:27
  • Okay. Take $n, n+1$ instead of $m, n$. Then use triangle inequality. – Jacky Chong Aug 05 '18 at 07:29
  • I'm sorry, I'm not convinced. – Bart Michels Aug 05 '18 at 07:45
  • Okay. Note that $|x_{i+1}-x_i| \leq \sum^{i-1}{k=0}\frac{1}{2^{k+1}}|y{i+1-k}-y_{i-k}|+\frac{1}{2^i}|x_2-x_1| \leq \sum^{i-1}_{k=0}\frac{\varepsilon}{2^{k+1}}+\frac{1}{2^i}|x_2-x_1| $ since ${y_n}$ is Cauchy. – Jacky Chong Aug 05 '18 at 07:53
  • I see two problems: that doesn't show yet that $x_n$ is Cauchy, since we we have no way to control $|x_2-x_1|$; also you estimated $|y_2-y_1|$ by $\epsilon$... – Bart Michels Aug 05 '18 at 08:01
  • It's a hint. Also, Cauchy is immediate by triangle inequality and geometric sum. – Jacky Chong Aug 05 '18 at 08:02
  • I see your concern, but we don't have to recursive back to $1$ and $2$. – Jacky Chong Aug 05 '18 at 08:07
  • Ok so I agree this gives $|x_{i+1} - x_i| \leq \epsilon$ for $i > N(\epsilon)$, but the triangle inequality gives you only $|x_n-x_m| \leq \epsilon \cdot |n-m|$... – Bart Michels Aug 05 '18 at 08:19
  • Okay. You are correct. I can't seem to find any elementary ways to salvage the idea. – Jacky Chong Aug 05 '18 at 08:47
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Hint. Call $y_n = 2x_{n+1} - x_{n}$, let $N>0$ and look at $$\frac1{2^N} y_n + \frac1{2^{N-1}}y_{n+1} + \cdots + \frac12 y_{n+N-1} + y_{n+N}$$

Bart Michels
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