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?

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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 $A_{n+1} \geq A_n$.

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

$$A_{n+1}=\sqrt{2+A_n} \le\sqrt{2+2}=2$$

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 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.

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