# 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?

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Nov 9 '18 at 0:11

## 4 Answers

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

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

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

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.