Wonderful sequences problem.

[daon]

Given any sequence {\(\displaystyle b_n\)} of natural numbers, we call the sequence {\(\displaystyle d_n\)}, where
\(\displaystyle \L d_n = a_{(n+b_n)} - a_n\) , a difference sequence.

Prove that the sequence {\(\displaystyle a_n\)} is Cauchy if and only if every difference sequence is a null sequence.
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I've spent some time figuring out what I'm supposed to do here, but I've got nothing really.

I assume the "null sequence" is the sequence having no elements, is that right? So

For the (=>) direction:
I assume {\(\displaystyle a_n\)} is Cauchy. Then for all \(\displaystyle \epsilon\) > 0 there exists \(\displaystyle N \in \mathbb{R}\) such that k, n > N implies that \(\displaystyle |a_k - a_n| < \epsilon\). I now don't know how to proceed. I also don't see how \(\displaystyle b_n\) ties into this at all.

For the (<= direction)
I assume {\(\displaystyle d_n\)} is the empty sequence. Then the sequence \(\displaystyle a_{(n+b_n)} - a_n\) is empty. Again, no idea where to go from here.

Thank you in advance,
-Daon

 

[pka]

No, a null sequence is one with zero as its limit.
That is an old and somewhat dated definition.

 

[pka]

Given any sequence {\(\displaystyle b_n\)} of natural numbers, we call the sequence {\(\displaystyle d_n\)}, where \(\displaystyle \L d_n = a_{(n+b_n)} - a_n\) , a difference sequence. Prove that the sequence {\(\displaystyle a_n\)} is Cauchy if and only if every difference sequence is a null sequence.
I assume {\(\displaystyle a_n\)} is Cauchy. Then for all \(\displaystyle \epsilon\) > 0 there exists \(\displaystyle N \in \mathbb{R}\) such that k, n > N implies that \(\displaystyle |a_k - a_n| < \epsilon\). I now don't know how to proceed. I also don't see how \(\displaystyle b_n\) ties into this at all.

Note that \(\displaystyle b_n \ge 0\) for all n because the \(\displaystyle b_n\) are natural numbers.

Thus if \(\displaystyle n \ge N\) then \(\displaystyle (n+b_n) \ge N\).

That is all one needs.

 

[daon]

Thank you pka. I somehow didn't make that connection. Using what you've pointed out, I've attempted it in two ways. I feel that the second one is more clear, but I could be wrong.

So, Since \(\displaystyle n>N \,\, \Rightarrow \,\, n+b_n > N\), I could say from the definition of Cauchy Criterion, {\(\displaystyle a_n\)} converges \(\displaystyle \Leftrightarrow \forall \elpsilon > 0\,\, |a_k-a_n|<\epsilon\) (or approaches zero, which is true since it is Cauchy). Then, since \(\displaystyle b_n + n\) > N, we have that \(\displaystyle |a_{b_n + n} - a_n|\) approaches zero.

{\(\displaystyle d_n\)} converges to zero \(\displaystyle \Leftrightarrow\) for all \(\displaystyle \epsilon\)>0 there is an N s.t. n>N => \(\displaystyle |a_{n+b_n}- a_n|<\epsilon\). From this definition we can extract a k=\(\displaystyle n+b_n\) (notice k>n>N) such that for all \(\displaystyle \epsilon\)>0, \(\displaystyle |a_k - a_n| < \epsilon\)

Thanks,
-Daon

 

[marcmtlca]

Suppose \(\displaystyle a_n\) is a Cauchy sequence. Therefore \(\displaystyle \exists N \forall m,n \geq N\) We have that \(\displaystyle |a_n-a_m|<\epsilon\) Noting as you did that \(\displaystyle n+bn \geq n\) It follows that \(\displaystyle |a_{n+b_n}-a_n|<\epsilon\) And therefore \(\displaystyle |d_n-0|<\epsilon\) Which says that \(\displaystyle d_n\) is a null sequence.

Suppose that \(\displaystyle d_n\) is a null sequence. Then \(\displaystyle \forall \epsilon \exists N \in \bf{N} \forall n\geq N\) We have that \(\displaystyle |d_n-0|<\epsilon\). This implies that \(\displaystyle |a_{n+b_n}-a_n-0|<\epsilon\) This is true independent of your choice of \(\displaystyle b_n\). It follows since \(\displaystyle b_n+n \geq n\) that \(\displaystyle a_n\) is Cauchy.