# Basic Questions About Inductive Sequences

Given a sequence $(X_n)$ defined as follows: $X_1>0$ and $\forall n, X_{n+1}=\frac{1}{2}(X_n+\frac{b}{X_n})$

what do I need to think about when I see the notion of $X_{n+1}$? Should I think $X_{n+1}$ as a sequence? or just a way to defined the next element of the sequence $X_n$? I'm asking this because I saw that the limit of $X_{n+1}$ is equal to the limit of $X_n$ and the limit is defined for sequences. Why are the limits equal? Can I talk about the sequence $X_n$ and $X_{n+1}$ interchangeably? If so, why?

Furthermore, in a lecture I saw involving the same sequences above, in order to find out if the sequence $X_n$ is decreasing we evaluted the expression $X_{n+1}-X_n$ which is equal to $\frac{-(X_n)^2+b}{2X_n}$ However, because we don't have a formula for $X_n$ he developed $\frac{-(X_{n+1})^2+b}{2X_n}$ instead and then he concluded that $\forall n, b-(X_{n+1})^2 \le 0$ from that he concluded that $\forall n \le2$, $b-(X_n)^2 \le 0$ why is that correct?

I'm quite confused - could you please help me to understand this fundamental concept?

Thank you very much for your time and help.

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First of all $X_{n+1}$ is not a sequence, it's a number. In particular it is the $(n+1)$'st term of the sequence $(X_n)$ (assuing that $(X_n)=(X_n)_{n=1}^\infty$).

The sequence $(X_{n+1})$ is indeed a sequence. Informally speaking, it is just the sequence $(X_n)$, but it starts with the term $X_2$, not $X_1$ (assuming again that $(X_n)=(X_n)_{n=1}^\infty$).

A concrete example:

If $\ \ \ (x_n)=(1,3,9,16,\ldots)$

then

$\ \ \ (x_{n+1})=(3,9,16,\ldots)$.

Obviously, these are different sequences, so you can't use "$(x_n)$" and "$(x_{n+1})$" interchangeably in a general statement. For example the statement "the second term of $(x_n)$/$(x_{n+1})$ is 3" would be false for the sequences defined above.

However, the two sequences $(y_n)$ and $(y_{n+1})$ will share many properties. For instance one is increasing if and only if the other is. Or, one has limit $L$ if and only if the other does. This is so because the two sequences are essentially the same, except that they have different "starting points".

With regards to your post: whenever you see $X_{n+1}$, you should regard it as a term of the sequence $(X_n)_{n=1}^\infty$ (the $(n+1)$'st term).

In particular the formula $$\tag{1} X_{n+1}={1\over2}\Bigl(X_n+{b\over X_n}\Bigr)$$ is giving you a relationship between the $(n+1)$'st and $n$'th terms of the sequence $(X_n)$.

Note my choice of using parentheses is sometimes unfortunate (it would probably be better to use braces for sequences: "${X_n}$").

For example in your line "then he concluded that $\forall n, b-(X_{n+1})^2 \le 0$ " you should regard "$(X_{n+1})$" as a term of the sequence $(X_n)$.

The only place in the argument where you consider the sequence $(X_{n+1})$ is when you argue that the sequences $(X_{n+1})$ and $(X_n)$ have the same limit. Then if you call this limit $L$, it follows from $(1)$ that $$L={1\over2}\Bigl(L+{b\over L}\Bigr).$$

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David, as always, thank you very very much for your great answers, I RALLY appreciate it! Now for a few questions: Firstly, how do I prove that the sequence $(X_{n+1})$ is the sequence $(X_n)$ that starts from $X_2$? Also, how do I prove that both $X_n$ and $X_{n+1}$ has the same limit? –  Anonymous Apr 5 '12 at 17:05
@Anonymous Look at the example carefully. For the sequence $(x_n)_{n=1}^\infty=(1,3,9,\ldots)$: For $n=1$, $x_{n+1}$ is just $x_2=3$; for $n=2$, $x_{n+1}$ is just $x_3=9$; and so on. So the sequence $(x_{n+1})_{n=1}^\infty=(x_2,x_3,\ldots)=(3,9,16,\ldots)$. You are just "shifting" the sequence $(x_n)_{n=1}^\infty$ to get the sequence $(x_{n+1})_{n=1}^\infty$. As far as having the same limit, this should be obvious as the limit of a sequence does not depend on what the first term of the sequence is. –  David Mitra Apr 5 '12 at 17:13
Is by definition when we write $(X_{n+1})$ we mean that we are copying the set $X_n$ from $X_2$? In other words is it a subsequence of $X_n$? Or in other words, just as we can talk about $X_{2n}$ which is taking all the even terms of $X_n$? Also how do I know when I'm taking about the term $X_(n+1)$ and the sequence $(X_{n+1})$? –  Anonymous Apr 5 '12 at 17:20
You need "$(X_n)$" not $X_n$ in your first and second sentences. Please be careful to distinguish a sequence $(X_n)$ from a term of the sequence $X_n$. But, yes, $(X_{n+1})$ is the subsequence of $(X_n)_{n=1}^\infty$ obtained by writing the terms of $(X_n)$ starting with $X_2$. –  David Mitra Apr 5 '12 at 17:24
OK, great. Also, when we write $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ we are talking about a new sequence that is defined by two sequences;1) the sequence ($a_n$) and 2) the sequence $(a_{n+1})$ which are two different sequences, correct right? –  Anonymous Apr 5 '12 at 17:27