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Given a sequence $(x_n)$ of real numbers satisfying $$|x_n - x_m| \leq \frac 12 |x_{n-1} - x_{m-1}|$$ does it follow that $(x_n)$ is Cauchy?

This question arose in an attempted answer to this question. I feel like this should be false. All I can think of is the double sequence $a_{n,m} = 2^{n-2m}$ which satisfies the analogous inequality above and does not go to $0$ for the product topology. But it does not satisfy the triangle inequality $a_{n,m} \leq a_{n,k} + a_{k,m}$ so it's quite far from a counterexample.

Bart Michels
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2 Answers2

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Yes.

Set $a=|x_1-x_0|$. By induction, applying your assumption with $n=m+1$, $$ |x_{m+1}-x_m| \le 2^{-m}a $$

This makes the sequence Cauchy: If you're given $\varepsilon$, choose $N$ large enough that $2^{-N}<\varepsilon/2a$. Then for every $n,m>N$ you have $|x_n-x_m|<2\cdot 2^{-N}a$.

hmakholm left over Monica
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Let $m=n-1$, then it is easy to show the series is Cauchy and convergent.

Or do you only have the condition for certain values of $n-m$?

Empy2
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