# Let $x_n$ be a real sequence. Suppose that there is an a>1 such that $|x_{n+1} -x_n|\le a^{-n}$.

Let $x_n$ be a real sequence.

Suppose that there is an $a>1$ such that $|x_{n+1} -x_n|\le a^{-n}$ for all $n\in\Bbb N$. Prove that $x_n \to x$ for some $x \in\Bbb R$.

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What have you tried? –  icurays1 Nov 30 '12 at 23:22

Since $a>1$, $\frac{1}{a} < 1$ and $\frac{1}{a^n} > \frac{1}{a^{n+1}}$. Then, for any $\epsilon > 0$ exists $n$ that makes your sequence a Cauchy sequence.