# limit of a sequence within a sequence

Suppose we have a sequence like this $A_0= 8$, $A_1=1$, $A_2=4$, $A_3=-1$, $A_4=2$, $A_5=1$, $A_6=1$, $A_7=-1$, $A_8=1/2,...$ (in other words the even terms get always divided by $2$, the odd terms alternate between $-1$ and $1$)

Can we somehow make sense of the idea that the subsequence of even terms approaches a limit? Also what if terms of a sequence indexed by prime numbers approach a limit, can we make sense of that?

• Sequenception.I don't understand the question though. Do you know the definition of subsequence? – Git Gud Dec 1 '13 at 20:31
• Well the question is: while it is clear that the sequence as a whole doesnt have a limit, can we make precise the intuitive idea that part of the sequence does approach a certain number? – Adam Dec 1 '13 at 20:33
• Yes, it is called a sublimit. A sublimit of a sequence is the limit of a subsequence of a sequence. – Git Gud Dec 1 '13 at 20:34
• Do you have a reference? – Adam Dec 1 '13 at 20:38
• It's in any book about sequences and probably any calculus/real analysis book. I don't think your question is a good fit for MSE, on account of its broadness. MSE is not a place to ask questions of the kind explain this to me. So I suggest you read my answer and if you don't have any questions, delete your question. – Git Gud Dec 1 '13 at 20:50

Given a sequence $(A_n)_{n\in \Bbb N}$ and a strictly increasing sequence of natural numbers $\alpha \colon \Bbb N\to \Bbb N$, the composition $A\circ \alpha$, often denoted by $(A_{\alpha _n})_{n\in \Bbb N}$, is a subsequence of $(A_n)_{n\in \Bbb N}$.

The limits of subsequences of $(A_n)_{n\in \Bbb N}$ are called sublimits of $(A_n)_{n\in \Bbb N}$.

In your question we have, for all $n\in \Bbb N$, $$A_n=\begin{cases}8, &\text{if }n=0\\\dfrac {8}{n}, &\text{if }n\text{ is even and }n\neq 0\\(-1)^{n+1}, &\text{if }n\text{ is odd} \end{cases}$$

You're thinking about the subsequence of the even terms of $(A_n)_{n\in \Bbb N}$. According to the definition above, why is that really a subsequence? Verify it is with $\alpha \colon\Bbb N\to \Bbb N, n\mapsto 2n$.

Since $\lim \limits_{n\to \infty}\left(A(\alpha(n))\right)=0$, it is true that $0$ is a sublimit of $(A_n)_{n\in \Bbb N}$.

• I was thinking about a following limit: let C(n) be nth composite number and LD(n) the smallest divisor of n greater than 1 and A(n)= n-LD(n)^2. What is the limit of A(C(n)) as n increases? – Adam Dec 1 '13 at 21:40
• I don't think that sequence even converges. On a separate note, I'd prefer you unnacepted my answer and complied with what I asked in a comment to the question. – Git Gud Dec 1 '13 at 21:44
• Please dont be offended, but you present after all an opinion of one person. I would rather wait for more people to express the same opinion. – Adam Dec 1 '13 at 21:47
• @Adam I'm not ofended, that's reasonable. – Git Gud Dec 1 '13 at 21:50