Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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


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

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 $$ L={1\over2}\Bigl(L+{b\over L}\Bigr). $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.