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Assume that $ C \in \mathbb{R}, q < 1$ and $|x_{n+1} - x_n | < Cq^n$ for all n. Prove that this is a Cauchy-sequence...

What I do is that I assume without loss of generality that $m < n$ so $n = m + k$.

Then by the triangle inequality, $$|x_{m+k} - x_m| \le |x_{m+k} - x_{m+k-1}| + |x_{m+k-1} - x_{m+k-2}| + \ldots + |x_{m-1} - x_m|$$ By using some formulas that we prove in a previous exercise, I get that above is less than $$\le C \sum_{i=m}^{m+k-1} q^i = C\frac{q^m - q^n}{1-q}$$ Is above (apart from this final step that you might not know of?) the correct method? And what more can I now do?

It's my first attempt at a epsilon proof for sequences, and I'm not sure if I can just say "pick N large so this stuff is less than epsilon", or if more is needed.

EDIT: q $\in (0,1)$

Marcus Ma
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1 Answers1

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You have done well and are almost there. Note that

$$\frac{q^m - q^n}{1-q} = q^m \frac{1-q^k}{1-q} \leq q^m\frac{1}{1-q} $$

for all $k \geq 1$. Define a new constant

$$ C' := \frac{C}{1-q}. $$

You have shown that for all $m \in \mathbb{N}$ and all $k \geq 1$ that

$$|x_{m+k} - x_{m}| \leq C' q^m. $$

Given $\epsilon > 0$ pick $N \in \mathbb{N}$ such that $q^N < \frac{\epsilon}{2C'}$. Then for all $m,n > N$ we have

$$|x_m-x_n| \leq |x_m - x_N| + |x_n - x_N| \leq 2C'q^N < \epsilon. $$

Frank
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