# Cauchy Sequence Example Clarification

I am working through an example that is helping me with a question. However, there is a step that I can't make sense of, so some clarification would be very helpful. I am refering to Example 6.5 from here. I'll paraphrase the example below:

Let $(x_n)$ be a sequence and suppose that

$$d(x_{n+1},x_{n+2}) \leq \frac{1}{2}d(x_{n+1},x_{n})$$

for all $n \geq 1$

Then $(x_n)$ is Cauchy. To show this, ﬁrst note that by assumption we have

$$d(x_{n},x_{n-1}) \leq \frac{1}{2^{n-2}} d(x_2,x_1)$$

so that if $n > m$ we have

$$d(x_m , x_n) \leq d(x_m,x_{m+1}) + \cdots + d(x_{n-1},x_{n})$$ $$\leq d(x_1 , x_2) \cdot \left [ \frac{1}{2^{m-1}} + \cdots + \frac{1}{2^{n-2}} \right ]$$ $$\leq d(x_1 , x_2) \cdot \left [ \frac{1}{2^{m-1}} + \frac{1}{2^{m}}+ \cdots \right ] = \frac{d(x_2 , x_1)}{2^{m-2}}$$ where we used the geometric formula for the last line.

I'll leave out the last part about how this shows that $(x_n)$ is Cauchy. I can't get to the same answer using the formula for a geometric series. I used

$$\sum_{k=m-1}^{n-2}=\sum_{k=0}^{n-2} 1/2^k - \sum_{k=0}^{m-2} 1/2^k$$

which, after applying the formula for a geometric series, gave me

$$\sum_{k=m-1}^{n-2}= \frac{1}{2^{m-2}} - \frac{1}{2^{n-2}}$$

which is obviously different from their answer. I then noticed that in the example

$$\left [ \frac{1}{2^{m-1}} + \cdots + \frac{1}{2^{n-2}} \right ]$$

becomes

$$\left [ \frac{1}{2^{m-1}} + \frac{1}{2^{m}}+ \cdots \right ].$$

What happened there? Also, just to clarify, $d(x_1, x_2) = |x_2 - x_1|$ ?

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They use that: $$\left [ \frac{1}{2^{m-1}} + \cdots + \frac{1}{2^{n-2}} \right ]\leq\left [ \frac{1}{2^{m-1}} + \frac{1}{2^{m}}+ \cdots \right ]=\\ =\dfrac{1}{2^{m-1}}\left [ 1+\frac{1}{2} + \frac{1}{2^2}+ \cdots \right ]=\dfrac{1}{2^{m-1}}\cdot2.$$