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I've always had the following silly(?) doubt about convergent sequences.

Given a finite sequence $A_{1:n}=(a_1,a_2\ldots ,a_n)$, define its reverse as $rev(A) = (a_n,a_{n-1},..a_1)$. Further, if $A$ is infinite, define its reverse as the limit of the reverses of its initial subsequences, i.e. as the limit of the sequence of sequences $ \{ (a_1), (a_2,a_1), (a_3,a_2,a_1), ..\}$.

Q: Is this notion well defined? Why or why not?

I suspect not, because the reverse of $(1,\frac{1}{2},\frac{1}{3},...)$ would be $(0,0,... , \frac{1}{3},\frac{1}{2},1)$? This seems absurd!

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    To put it another way, the concepts of "last" and "infinite" are mutually exclusive.2010-12-17

1 Answers 1

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This "reverse sequence" notion is either going to be meaningless, or be identical to sequences (but just written backwards). For it to be well-defined, you need somehow to index the entries.

As it is defined, the sequence of sequences $(a_1)$, $(a_2,a_1)$, etc., would converge pointwise to the sequence $(\ell,\ell,\ldots)$, where $\ell = \lim_{n \rightarrow \infty} a_n$ (if it exists). So this is not particularly exciting.

The problem with the $(\ell,\ell,\ldots,a_N,a_{N-1},\ldots,a_1)$ idea, is that you need to take $\infty-\infty=N$ (which will give rise to nonsense).

If you defined the sequence $\mathrm{rev}(A)$ such that the $(-i)$-th entry of $\mathrm{rev}(A)$ is the $i$-th entry of $A$, then it will also be well-defined, but identical to $A$ itself (just written backwards). So this is also not particularly exciting.