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