How to prove the sequence $X_n=(1+X_{n-1})/2$ is bounded-above or -below given $X_0$ I'm stuck on this step for my homework of the class of real analysis. The question says that suppose $X_0 \in R , X_n = (1+X_{n-1})/2 \: \forall n \in N$, Use Monotone Convergence Theorem to prove $X_n \to 1 \: as \: n \to \infty$. 
I used the subtraction method between $X_{n+2}$ and $X_{n+1}$ and have found that this sequence is monotone in 3 different cases. But since M.C.T requires it to be monotone and bounded-above or -below, I then have been having difficulty in proving so for this sequence. It seems to me that regardless being increasing or decreasing, this sequence is not bounded at all. I've asked in the inner forum for my class and the teacher but haven't got anything so far. Any hints are very appreciated! Thanks!!
Thanks for the answers of changing the expression to find its limit is 1. But the question asks specifically to use MCT. So my understanding is that I need to prove the sequence fulfills the assumptions of MCT first. Then I need to prove the supremum of this sequence is 1. So finally I can say the statement is proved to be true by MCT, which says the limit of an increasing and bounded-above sequence is its supremum.
 A: Another observation:
$$X_n=\frac{1}{2}+\frac{X_{n-1}}{2}=\frac{1}{2}+\frac{1}{4}+\frac{X_{n-2}}{4}=
\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{n}}+\frac{X_0}{2^{n}}=\\
\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n-1}}\right)+\frac{X_0}{2^{n}}=\\
\frac{1}{2}\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)+\frac{X_0}{2^{n}}=1-\frac{1}{2^n}+\frac{X_0}{2^{n}}\rightarrow1, n\rightarrow\infty$$
A: $x_n=1/2(x_{n-1}+1)\implies x_n-1=1/2(x_{n-1}-1)\implies x_n-1=\frac{x_0-1}{2^n}\rightarrow 0.$ Hence $x_n\rightarrow 1.$
A: Geometric intuition : the new value is the midpoint between $1$ (which is fixed) and the previous value. Thus, the distance to $1$ is halved at each iteration...  Thus the distance to $1$ tends to $0$... meaning that $X_k$ is "attracted" by $1$, therefore the limit point of the sequence. 
A: 
It seems to me that regardless being increasing or decreasing, this sequence is not bounded at all. 

This is incorrect.  Did you try checking what the sequence looks like for some value of $X_0$?
