prove this sequence is increasing The sequence $A_n$ is described as-
\begin{align*}
A_1 & = \sqrt{2}\\
A_{n+1} & =(2+A_n)^{0.5}
\end{align*} 
where $A_n$ is the $n$th term in the sequence.
I can prove by induction that this sequence is bounded by above by $2$ and then find the limit to be $2$. But in order to assume that the limit exists, I need to show that it is also increasing so I can use the monotone convergence theorem. It is easy to see why it is increasing, but how do I prove it?
 A: We need to prove that
$$A_{n+1}=\sqrt{2+A_n}\ge A_n \iff A_n^2-A_n-2 \le 0 \iff -1\le A_n \le2$$
then we have $A_n \ge 0$ and by inducion we can prove that $A_n\le 2$ indeed


*

*base case: $A_1=\sqrt 2\le 2$

*induction step: assuming true $A_n\le 2$ we need to prove that $A_{n+1}\le 2$ then


$$A_{n+1}=\sqrt{2+A_n} \le\sqrt{2+2}=2$$
A: Prove by induction that $A_n \leq 2$. Then note that $(A_n-\frac  1 2)^{2} \leq \frac 9 4$. When expanded this reduces to $A_{n+1} \geq A_n$.
A: If we can assume $A_n, A_{n+1} > 0$ then 
$A_{n+1} > A_n \iff A_{n+1}^2 > A_n^2$
And $A_{n+1}^2n = ((2 + A_n)^{\frac 12}) = 2 + A_n$
If we can assume $A_n < 2$ then $A_{n+1}^2 = 2 + A_n > A_n + A_n > 2*A_n > A_n*A_n = A_n^2$.
And that's that.
.....
We do have to prove that $A_n > 0$ but that's trivial:
By induction: Base case: $\sqrt 2 > 0$.  Induction step.  If $A_n >0$ then  $A_n + 2 > 0$ so $A_{n+1} > \sqrt{A_n + 2} > 0$.
And we have to prove $A_n < 2$  but that's easy:
By induction: Base case: $\sqrt 2 <2$.  Induction step. If $A_n < 2$ then $A_n + 2 < 4$ and $A_{n+1} = \sqrt{2+A_n} < \sqrt {4} = 2$.
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If we start from the begining we can to it is one fell swoop:
Proposition:  $0 < A_1 < A_2 < .... A_n < A_{n+1} <.... <  2$.
Base Case:  $0 < A_1 = \sqrt 2 < 2$
Induction case: Suppose $0  < A_n < 2$ then
$0 < A_n^2 < 2A_n=A_n + A_n < A_n + 2 < 2+2=4$
so $0< \sqrt{A_n^2} < \sqrt{A_{n + 2}} < \sqrt{4}$
and $0< A_n < A_{n+1} < 2$.
A: A proof by contradiction.
Suppose to the contrary that for some term, we have $A_{k+1} \leq A_k$.  Suppose that this is the first instance for which this occurs.  Now, we have that 
$$A_{k+1} = \sqrt{2+A_k},$$ so that by using our assumption,
$$ 2 + A_k= A_{k+1}^2  \leq A_k^2.$$
Now, since $$A_k = \sqrt{2+A_{k-1}},$$
then also $$A_{k}^2 = 2+A_{k-1}.$$
So, plugging in above, we have that 
$$2+A_k \leq A_{k}^2 = 2+A_{k-1}.$$
In other words, $$A_k \leq A_{k-1}.$$
This is a contradiction to the assumption that $A_{k+1}$ was the first instance for which this occurs.  Thus, the sequence must be increasing.  
