How to show $c_n \leq c_{n+1}$?

Question

How to show $c_n \leq c_{n+1}$?

Let $c_1=\sqrt{2}$ and $c_{n+1}=\sqrt{2+c_n}$. I have already proven that $c_n \leq 1+\sqrt{2}$.

If I try to prove that $c_n \leq c_{n+1}$, I end up by showing that it does not work:$$c_n\leq 1+\sqrt{2} \leq c_{n+1}=\sqrt{2+c_n}$$$$\Leftrightarrow c_n^2 \leq 2+c_n$$ This is not true, what am I doing wrong?

Answer 1

If $c_n\le c_{n+1}$, then $$c_{n+1}=\sqrt{2+c_n}\le\sqrt{2+c_{n+1}}=c_{n+2}$$ etc.

Answer 2

• It is easy to see by induction that $c_n\leq 2$ for all $n$: $$c_{n+1}\leq \sqrt{2+2} = 2$$
• Clearly all elements are positive, so $$c_{n+1}\geq c_n\iff c_{n+1}^2\geq c_n^2 \iff 2+c_n \geq c_n^2$$

$$\iff 0\geq (c_n-2)(c_n+1)\iff c_n\leq 2$$ which is true.

Answer 3

If $c_n \le 1 + \sqrt 2$ for all $n$ then $c_{n+1} \le 1 + \sqrt 2$ as well so you we never be able to prove $c_n \le 1+ \sqrt 2 < c_{n+1}$ because it simply is not true. $1+\sqrt 2$ serves as an upper limit and can be quite a bit more larger than $c_n$ than $c_{n+1}$ is.

(Just like if you want to prove $6 < 8$ and you already proved $6 < 5,127$ it won't work to think "Okay I just need to prove $6 < 5,127 < 8$".)

You will need to compare $c_n$ to $c_{n+1}$ directly.

Any way this just cries out for a proof by induction.

If we assume we know that $c_{n-1} < c_{n}$ we can use that to prove:

$2 + c_{n-1}< 2 + c_{n}$ and if we can assume $c_{n-1} \ge 0$ we have.

$0 \le \sqrt {2 + c_{n-1}} < \sqrt{2 + c_n}$ and

$0 \le c_n < c_{n+1}$.

That would be a perfect induction step.

We just need a base step.

We need to prove $0 \le c_1 < c_2$

$c_2 = \sqrt {2+ \sqrt 2} > \sqrt{2} = c_1$ and $\sqrt{2} \ge 0$ so that's that.

Answer 4

Note that $$ \begin{align} c_{n+1}-2 &=\sqrt{2+c_n}-2\\ &=\frac{c_n-2}{\sqrt{2+c_n}+2}\tag1 \end{align} $$ Equation $(1)$ says that $c_{n+1}-2$ and $c_n-2$ have the same sign, and that $$ |c_{n+1}-2|\le\frac{|c_n-2|}2\tag2 $$

That is, when $c_n\lt2$, we have that $c_n\lt c_{n+1}\lt2$. This is the case when $c_1=\sqrt2$ .