I am reading a book (Real Analysis 1 by Terence Tao), where the author presents a proof that $\lim_{n \rightarrow \infty} x^{n} = 0$ for $0 < x < 1$. In the middle of the proof, he says following:
…… But the sequence $(x^{n+1})_{i=1}^{\infty}$ is just the sequence $(x^{n})_{n=2}^{\infty}$ shifted by one, and so they must have the same limits ……
If one writes down the first elements of both sequences, it becomes obvious that these two sequences are indeed "the same":
$$(x^{2},x^{3},x^{4},\cdots)$$
However, these two sequences are not equal (unless I am misunderstanding the definition of the sequences). By definition (the one presented in the book), a sequence $(a_{i})_{i=m}^{\infty}$ is defined as a function $f: \{n \mid n ≥ m\} \rightarrow \mathbb R$, where $f(i) = a_i$. Using this definition, one can see that the sequences $(x^{n+1})_{n=1}^{\infty}$, $(x^{n})_{n=2}^{\infty}$ cannot be equal, since the function corresponding to the sequence $(x^{n+1})_{i=1}^{\infty}$ has $1$ in its domain, but the function corresponding to $(x^{n})_{n=2}^{\infty}$ does not (i.e., the underlying functions have different domains, hence they are not equal, which means the sequences are not equal).
In fact, even if one takes a simple sequence $(1,2,3,\cdots)$, one can find infinitely many ways to denote such sequence (e.g. $(n)_{n=1}^{\infty}$, $(n-1)_{n=2}^{\infty}$, $(n-2)_{n=3}^{\infty}$ etc.), and according to our definition, all of those denotations refer to the different sequences (i.e., $(n)_{n=1}^{\infty} ≠ (n-1)_{n=2}^{\infty}$, $(n)_{n=1}^{\infty} ≠ (n-2)_{n=3}^{\infty}$ etc.)
Question.
Is there an alternative definition that would allow us to equalize sequences like $(n)_{n=1}^{\infty}$, $(n-1)_{n=2}^{\infty}$, $(n-2)_{n=3}^{\infty}$ or sequences like $(x^{n+1})_{i=1}^{\infty}$, $(x^{n})_{i=2}^{\infty}$?
There is one particular reason why I think this alternative definition could be useful. Let's consider an example:
Let $0 < x < 1$. Also let's assume that we've already shown that the sequence $(x^{n})_{i=1}^{\infty}$ is convergent to $L$. It is clear that sequences $(x^{n-1})_{i=2}^{\infty}$,$(x^{n-2})_{i=3}^{\infty}$, $(x^{n-3})_{i=4}^{\infty}$ $\cdots$ are the "same" as $(x^{n})_{i=1}^{\infty}$, in a sense that they all correspond to the sequence
$$(x^{2},x^{3},x^{4},\cdots)$$
And hence is it quite reasonable to conclude that those sequences must also converge to $L$. However, since all those sequences are not equal to each other, without rigorous definition of "sameness", the claim should be justified for each sequence separately. For example, if we want to show that $(x^{n-1})_{i=2}^{\infty}$ also converges to $L$, we need to take $\epsilon > 0$ and find an $N$ such that for all $n ≥ N$, $|a^{n-1} – L| ≤ \epsilon$ (it is easy to do so, of course, but is rather cumbersome; what's worse, we still need to show the convergence for the sequences $(x^{n-2})_{i=3}^{\infty}$, $(x^{n-3})_{i=4}^{\infty}$). But if one can come with a rigorous definition of "sameness", the notion of convergence of the "same" sequences can be generalized.
Best Answer
As suggested in a comment, you could define $(a_i)_{i=m}^\infty$ as the function $f:\Bbb N\to\Bbb R$, $f(i)=a_{m-1+i}$.