If ##|s_{n+1} - s_n| \lt 1/2^n##, then ##(s_n)## is a Cauchy sequence

In summary, it can be proved that for any given ##\varepsilon > 0##, there exists an ##N## such that ##n > N## implies ##\frac{1}{2^n} < \varepsilon##. By using the triangle inequality repeatedly, we can show that the sequence ##(s_n)## is a Cauchy sequence and therefore converges to a limit.
  • #1
Hall
351
88
Homework Statement
No entry
Relevant Equations
No entry
My attempt: It can be proved that ##\lim \frac{1}{2^n} = 0##. Consider, ##\frac{\varepsilon}{k} \gt 0##, there exists ##N##, such that
$$
n \gt N \implies \frac{1}{2^n} \lt \varepsilon
$$
Take any ##m,n \gt N##, and such that ##m - k = n##.
##|s_m - s_{m-1} | \lt \frac{1}{2^{m-1}} \lt \frac{\varepsilon}{k}##
##| s_{m-1} - s_{m-2} | \lt \frac{1}{2^{m-2}} \lt \frac{\varepsilon}{k}##
##| s_{m-2} - s_{m-3} \ \lt \frac{1}{2^{m-3} } \lt \frac{\varepsilon}{k}##
## \vdots##
##| s_{m-k+1} - s_{m-k} | \lt \frac{1}{2^{m-k} } \lt\frac{\varepsilon}{k}##
By using triangle inequality repeatedly, we have
## | s_m - s_{m-k} | \lt \varepsilon##
##| s_m - s_n| \lt \varepsilon##

But my doubt with this method is that my choice of ##\frac{\varepsilon}{k}## was not, I mean, very arbitrary, that ##k## determined ##m## and ##n##.

And why everyone else on internet is doing it by using the triangle inequality and making it less than ##\sum_{k=n}^{m-1} \frac{1}{2^k}## and not to ##\varepsilon##?
 
Physics news on Phys.org
  • #2
What about ##m = n +k +1##?
 
  • #3
The larger you make/consider your tail, the larger the differences, which are then not neccesarily staying within ##\epsilon## from each other. Maybe using ##|S_{n+1} -S_{n}| < \epsilon/2^{n} ## could work.
 
Last edited:
  • #4
WWGD said:
The larger you make/consider your tail, the larger the differences, which are then not neccesarily staying within ##\epsilon## from each other. Maybe using ##|S_{n+1} -S_{n}| < \epsilon/2^{n} ## could work.
Can you please explain a little more?
 
  • #5
The worst case for you is if the sequence is strictly increasing, and each ##s_{n+k}## tries to get as far away from ##s_n## as possible. How far away can it go?
 
  • #6
Office_Shredder said:
The worst case for you is if the sequence is strictly increasing, and each ##s_{n+k}## tries to get as far away from ##s_n## as possible. How far away can it go?
##\sum_{j=n}^{n+k-1} \frac{1}{2^j}##
 
  • #7
Can you write down an upper bound for that (let k go to infinity)?
Then use that to say something about the sequence being cauchy.
 
  • #8
Office_Shredder said:
Can you write down an upper bound for that (let k go to infinity)?
Then use that to say something about the sequence being cauchy.
The maximum distance between any two elements of sequence ##(s_n)##, can be found by letting ##k \to \infty## in the sum above (which is the same thing to say that the distance between ##s_n## and a very far off term), so, we have (the notation ##s_{\infty}## might not be very okay)
$$
| s_{\infty} - s_n| \lt \sum_{n}^{\infty} \frac{1}{2^j}$$
$$
| s_{\infty} - s_n| \lt \frac{1}{2^{n-1} }$$

$$
\because \lim \frac{1}{2^n} = 0$$
$$
\therefore n \gt N \implies \frac{1}{2^n} \lt \varepsilon$$

Take any ##m,n \gt N+1 ##, we have established above that no matter how far away ##m## is from ##n##, we have an upper bound, so,
$$
| s_m - s_n| \lt \frac{1}{2^{n-1} }$$
As ## n \gt N+1##, the least value ##n## can take is ##N+2##, therefore RHS of the above inequality would become ## \frac{1}{2^{N+1} }## which is, of course, is less that ##\epsilon##.
$$
| s_m - s_n| \lt \varepsilon $$
 
  • #9
I think you have the right idea, but it's not well written. The first line should probably look like: given ##\epsilon##, pick ##N## such that ##\frac{1}{2^N} < \epsilon##
 
  • Like
Likes malawi_glenn
  • #10
Office_Shredder said:
I think you have the right idea, but it's not well written. The first line should probably look like: given ##\epsilon##, pick ##N## such that ##\frac{1}{2^N} < \epsilon##
Let me try to re-write it.

For a given ##\varepsilon \gt 0##, there exists ##N## such that ## n \gt N \implies \frac{1}{2^n} \lt \varepsilon##.

Take any ##m, n \gt N+1##, and we can safely assume ##m-n =k## for some positive ##k##. So, we have

##|s_m - s_{m-1} | \lt \frac{1}{2^{m-1}}##
##| s_{m-1} -s_{m-2}| \lt \frac{1}{2^{m-2} } ##
##\vdots##
##|s_{m-k+1} - s_n| \lt \frac{1}{2^n}##
By using triangle inequality repeatedly, we have
##| s_m - s_n| \lt \sum_{j= n}^{m-1} \frac{1}{2^j} \lt \sum_{n}^{\infty} \frac{1}{2^j}##
##|s_m - s_n| \lt \frac{1}{2^{n-1} } \lt \varepsilon##.

Hence, ##(s_n)## is a Cauchy sequence.
 
  • Like
Likes malawi_glenn

FAQ: If ##|s_{n+1} - s_n| \lt 1/2^n##, then ##(s_n)## is a Cauchy sequence

1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers where the terms get closer and closer together as the sequence progresses. In other words, for any given small number, there is a point in the sequence where all subsequent terms are within that small number of each other.

2. Why is the condition ##|s_{n+1} - s_n| \lt 1/2^n## important for proving that a sequence is Cauchy?

This condition ensures that the terms in the sequence are getting closer together at a faster rate compared to the terms in the denominator. This is necessary to show that the sequence is eventually contained within any given small number, which is a key characteristic of a Cauchy sequence.

3. How is the condition ##|s_{n+1} - s_n| \lt 1/2^n## related to the definition of a Cauchy sequence?

The condition is directly related to the definition of a Cauchy sequence, which states that for any given small number, there is a point in the sequence where all subsequent terms are within that small number of each other. The condition ensures that the terms in the sequence are getting closer together at a faster rate, making this definition possible.

4. Is the condition ##|s_{n+1} - s_n| \lt 1/2^n## sufficient for proving that a sequence is Cauchy?

Yes, the condition is sufficient for proving that a sequence is Cauchy. However, there are other conditions that can also be used to prove the Cauchy-ness of a sequence, such as the Cauchy convergence criterion or the Bolzano-Weierstrass theorem.

5. Can a sequence be Cauchy without satisfying the condition ##|s_{n+1} - s_n| \lt 1/2^n##?

Yes, a sequence can be Cauchy without satisfying the condition ##|s_{n+1} - s_n| \lt 1/2^n##. This condition is just one of several ways to prove that a sequence is Cauchy, but it is not the only way. As long as the sequence satisfies the definition of a Cauchy sequence, it can be considered Cauchy.

Back
Top