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If $\{x_n\}_n$ is a sequence of real numbers define as follow: $$|x_{n+1}-x_n|<r^{n-1}$$for all integer $n\geq1$ and $0<r<1$. Show that it's a Cauchy sequence.


We know that If the sequence $\{x_n\}_n$ converges, then it is Cauchy.

I'm thinking about using the squeeze theorem but the problem is that we have a strict inequality. Im not even sure if we are allowed to use the squeeze theorem on recurrent sequence.

\begin{array}{cc} &|x_{n+1}-x_n|<r^{n-1}\\ \implies & -(r^{n-1})<x_{n+1}-x_n<r^{n-1} \\ \implies& \lim_{n\to\infty} -(r^{n-1}) <\lim_{n\to\infty} x_{n+1}-x_n < \lim_{n\to\infty} r^{n-1}\\ \implies & 0< \lim_{n\to\infty} x_{n+1}-x_n < 0 \end{array}

First, I wanna know in general if can we use the squeeze theorem on strict inequality like maybe substracting a small $\epsilon$ so that it can $\leq$.

Secondly, can we use the squeeze theorem on recurrent sequence.

Thank you.

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For $m>n$ $$ |x_m-x_n|=|x_m-x_{m-1}+x_{m-1}-x_{m-2}+\cdots-x_n|\le\sum_{j=n}^{m-1}|x_{j+1}-x_{j}|<\sum_{j=n}^{m-1}r^{j-1}\\=r^{n-1}{1-r^{m-n}\over1-r}<{r^{n-1}\over1-r} $$ Given $\varepsilon>0\exists k:n>k\implies {r^{k-1}\over1-r}<\varepsilon$ as $$\lim_{n\to\infty}{r^{n-1}\over1-r}=0$$

You can use squeeze theorem on recurrent sequences and limit operations turn strict inequalities into weak inequalities.

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