Need help proving sequence is cauchy sequence and hence convergent.
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1st line: Re-writing as a telescopic sum.
2nd line: Triangle inequality
3rd line: Follows from definition shown in post #1. For example, for the first difference, let m = n + 1, which implies n = m - 1, so 2^-n = 2^(1-m)
4th line: Factor out 2^-n
5th line: Geometric series
6th line: Simplify
7th line: Finish showing it's Cauchy by using definition of Cauchy sequence.
EDIT:// Let me know if more detail is needed. I want to keep details light to get a better grasp for which steps labelled here need more explanation.
2nd line: Triangle inequality
3rd line: Follows from definition shown in post #1. For example, for the first difference, let m = n + 1, which implies n = m - 1, so 2^-n = 2^(1-m)
4th line: Factor out 2^-n
5th line: Geometric series
6th line: Simplify
7th line: Finish showing it's Cauchy by using definition of Cauchy sequence.
EDIT:// Let me know if more detail is needed. I want to keep details light to get a better grasp for which steps labelled here need more explanation.
Last edited:
m, n and are labels only. Choice to make m > n seems usual based on textbook definition of Cauchy. Cauchy sequence - Wikipedia
For line 5, notice that there is a strict less than sign.
This is because in line 4, we have a sum of finite terms with r = 1/2. Writing the terms in the brackets in reverse we see the sum from n=0 (1 = (1/2)^0) to (1/2)^(m-n+1)), and since m>n, m-n+1 is a positive finite number.
This sum of finite terms is strictly less than the sum of infinite terms shown in the brackets on line 5, computed as 1/(1-(1/2)) = 2
For line 5, notice that there is a strict less than sign.
This is because in line 4, we have a sum of finite terms with r = 1/2. Writing the terms in the brackets in reverse we see the sum from n=0 (1 = (1/2)^0) to (1/2)^(m-n+1)), and since m>n, m-n+1 is a positive finite number.
This sum of finite terms is strictly less than the sum of infinite terms shown in the brackets on line 5, computed as 1/(1-(1/2)) = 2
Assuming that \(\displaystyle m>n\) can be done without the lost of any generality(i.e. it does no harm).
Many of us simply I no idea what you know about sequences & series. Therefore, makes giving you help very difficult.
Here are sone facts: if the series \(\displaystyle \sum\limits_{k = 1}^\infty {{S_k}} \) converges and \(\displaystyle c>0\) then there exists a positive integer \(\displaystyle N\) such \(\displaystyle \left| {\sum\limits_{k = N}^\infty {{S_k}} } \right| < c\).
That tells us that the "tail-end" of a convergent series is small. So any part of the tail is very small.
That is the driving concept behind a Cauchy sequence.
Now what we have is \(\displaystyle \left| {\sum\limits_{k = N}^\infty {{2^{-k}}} } \right| \) is convergent.
Can turn that into a Cauchy sequence ?