Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, $d(x_1, x_2) = |x_2 - x_1|$ ?

share|improve this question

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.