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Cauchy Sequences

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Hello, as you can see, this proof is about why sqrtnsqrt{n} is not a Cauchy sequence. According to my definition, I know what the negation of the definition is. I have two questions:

- Here we choose an ε and we also choose m and n. My question is: We need that m and n > N still, so do we choose our 'm' and 'n' based on N right? E.g. m and n have to be functions of N? Here, I guess they chose to use 'N' as 'n'. But, I mean that I'd do |am - an| to see it is equal to |sqrt m - sqrt n| and then i'd see what two functions of N can I choose such that I get this to be greater than or equal to a number e.g. if I chose epsilon = 4, I'd need that m = 25n... where m and N are both functions of N?

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Reply 1
The game is (restating definition) that for every epsilon, there exists an N such that every n,m > N, then the difference of the sequence terms is smaller than epsilon.

The sqrt sequence breaks this as for any N, we an choose n=N and m=4N=4n (here were being a bit lax in that n=N but that's not important). The difference is sqrt(n)=sqrt(N). This will not be smaller than any given epsilon as we can make the difference as large as necessary by choosing N sufficiently large.

The sqrt sequence tends to infinity although the difference between successive terms goes to zero. Choosing m=4n could have been m=9n or ... and the difference can be made as large as you want for any N.
Original post by mqb2766
The game is (restating definition) that for every epsilon, there exists an N such that every n,m > N, then the difference of the sequence terms is smaller than epsilon.

The sqrt sequence breaks this as for any N, we an choose n=N and m=4N=4n (here were being a bit lax in that n=N but that's not important). The difference is sqrt(n)=sqrt(N). This will not be smaller than any given epsilon as we can make the difference as large as necessary by choosing N sufficiently large.

The sqrt sequence tends to infinity although the difference between successive terms goes to zero. Choosing m=4n could have been m=9n or ... and the difference can be made as large as you want for any N.

Ok, thank you. So, it seems like we don't choose an 'N' here, it is arbitrary and it has to work for all possible values of n. Instead, we choose our 'm' and 'n' based on this arbitrary N right such that the negation of the definition is satisfied? Oh right, so the chosen value of n was N. Oh yes and as n/N is a term of the sequence, i mean not the actual term, but the 'nth' term or whatever, then it has to be a natural number, so n >= 1, hence sqrt n >= 1 by that.

Oh yes, I see thank you. So, I could've, to fix everything up, chosen:
epsilon = 2, m = 16N, n = 4N, so |am - an| = |sqrt(16N) - sqrt(4N)| = 4sqrt(N)-2sqrt(N) = 2sqrt(N) >= 2 as sqrt(N) >= 1 because N>=1.
Reply 3
Original post by Takeover Season
Ok, thank you. So, it seems like we don't choose an 'N' here, it is arbitrary and it has to work for all possible values of n. Instead, we choose our 'm' and 'n' based on this arbitrary N right such that the negation of the definition is satisfied? Oh right, so the chosen value of n was N. Oh yes and as n/N is a term of the sequence, i mean not the actual term, but the 'nth' term or whatever, then it has to be a natural number, so n >= 1, hence sqrt n >= 1 by that.

Oh yes, I see thank you. So, I could've, to fix everything up, chosen:
epsilon = 2, m = 16N, n = 4N, so |am - an| = |sqrt(16N) - sqrt(4N)| = 4sqrt(N)-2sqrt(N) = 2sqrt(N) >= 2 as sqrt(N) >= 1 because N>=1.


Sure, if you prove that the m-n difference can be arbitrarily large for any N, it can't be Cauchy.The
(edited 3 years ago)
Original post by mqb2766
Sure, if you prove that the m-n difference can be arbitrarily large for any N, it can't be Cauchy.

Thank you!

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Just one more thing, I am really confused on a subsequence and how to interpret it. What is 'n_k' in this case? Can I think of it like n_k is just any strictly increasing function of n, e.g. n_k = 2n, n_k = n^2 etc... and of course without all of that, if we have the sequence a1, a2, a3, ..., I can choose the subsequence to be a1, a5, a9 etc.. and is this a subsequence because we choose our 'n_k' here as 1, 5, 9 which are strictly increasing?

Also, in that case, the terms of a subsequence have to be a 'subset' of the terms of the sequence right i.e. every term in the subsequence must also be at least one of the terms of the sequence? Like they are particular terms of the original sequence but I guess skipping terms sort of and making sure you have no consecutive terms basically?
(edited 3 years ago)
Reply 5
Original post by Takeover Season
Thank you!

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Just one more thing, I am really confused on a subsequence and how to interpret it. What is 'n_k' in this case? Can I think of it like n_k is just any strictly increasing function of n, e.g. n_k = 2n, n_k = n^2 etc... and of course without all of that, if we have the sequence a1, a2, a3, ..., I can choose the subsequence to be a1, a5, a9 etc.. and is this a subsequence because we choose our 'n_k' here as 1, 5, 9 which are strictly increasing?

Also, in that case, the terms of a subsequence have to be a 'subset' of the terms of the sequence right i.e. every term in the subsequence must also be at least one of the terms of the sequence? Like they are particular terms of the original sequence but I guess skipping terms sort of and making sure you have no consecutive terms basically?

The terms could be consecutive as a sequence could be a subsequence of itself n_k = k. But basically yes.
Original post by mqb2766
The terms could be consecutive as a sequence could be a subsequence of itself n_k = k. But basically yes.

Oh right yes, of course, taking the natural numbers itself is increasing as well. So, you are just saying one subsequence of a sequence is also just the sequence itself?
Reply 7
Original post by Takeover Season
Oh right yes, of course, taking the natural numbers itself is increasing as well. So, you are just saying one subsequence of a sequence is also just the sequence itself?

Using the definition you gave, yes.
Original post by mqb2766
Using the definition you gave, yes.

Thank you, perfect!

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