# Show that $s_n<2$

If $s_1=\sqrt{2}$ and $s_{n+1}=\sqrt {2+s_n}$ for $n\geq 1$ show that $s_n<2$ $\forall, n\geq1$ and $s_n$ is convergent.

The second part I supposed that is immediate from Cauchy sequence definition that $s_n$ is convergent.

The first part I think to use induction

For $n=1$ we have that $s_1=\sqrt{2}<2$

Suppose that is true for $n=k$ then we need to show for $n=k+1$. I wrote a few terms of this recursive sequence

$$s_1=\sqrt{2},\quad s_2=\sqrt{2+\sqrt{2}},\quad s_3=\sqrt{2+s_2}=\sqrt{2+\sqrt{2+\sqrt{2}}}$$ Then I think that $s_{n+1}$ can written as $$s_{n+1}=\sum_{i=1}^n2^{\frac{1}{2n}}$$

I do not know how to argue that the above term is less than $2$.

• $$s_{n+1} = \sqrt{2+s_n} < \sqrt{2 + 2} = 2$$ – peterwhy Jun 5 '17 at 23:25
• Your general form is wrong. – Simply Beautiful Art Jun 5 '17 at 23:25
• And it is not "obvious" that the sequence is Cauchy – Luiz Cordeiro Jun 5 '17 at 23:28

## 5 Answers

As peterwhy mentions,

$$s_n<2\implies s_{n+1}=\sqrt{2+s_n}<\sqrt{2+2}=2\tag{\color{green}\checkmark}$$

To show that it is convergent, you want to show that it is monotone increasing, which combined with the fact that it is bounded above will mean it converges:

$$s_n>s_{n-1}\implies s_{n+1}=\sqrt{2+s_n}>\sqrt{2+s_{n-1}}=s_n\tag{\color{green}\checkmark}$$

and thus you are done.

By the induction assumption, for some $k\in\mathbb N$, $$s_k < 2$$

Consider $n = k+1$,

$$s_{k+1} = \sqrt{2+s_k} < \sqrt{2 + 2} = 2$$

Together with the fact that $s_1 = \sqrt 2 < \sqrt 4 = 2$, by induction, all $s_n < 2$.

For $0<x<2$, $x^2 < 2x$ and $2x < 2 + x$.

To prove that the sequence $s_n$ is increasing,

\begin{align*}s_{n+1} &= \sqrt{2+s_n}\\ &> \sqrt{s_n + s_n}\\ &= \sqrt{2s_n}\\ &> \sqrt{s_n^2}\\ &= s_n \end{align*}

Since the sequence $s_n$ is increasing and bounded above, limit exists.

Hint: $\,s_{n+1}-2=(\sqrt {2+s_n}-2) \cdot \cfrac{\sqrt {2+s_n}+2}{\sqrt {2+s_n}+2} = \cfrac{2+s_n-4}{\sqrt {2+s_n}+2} = \cfrac{s_n-2}{\sqrt {2+s_n}+2}\,$, therefore $s_{n+1}-2$ has the same sign as $s_n -2\,$ and, by induction/telescoping, the same sign as $s_1-2\,$.

Using the fact that the square root function is increasing, and the induction hypothesis that $s_k<2$, we have: $$s_{k+1}=\sqrt{2+s_k}<\sqrt{2+2}=2$$

Assuming that the sequence does converge (you prove that later) write the limit as "S". Letting n go to infinity in $s_{n+1}= \sqrt{2+ s_n}$ we get $S= \sqrt{2+ S}$. Squaring both sides, $S^2= 2+ S$ so S must satisfy $S^2- S- 2= (S- 2)(S+ 1)= 0$ so S is either 2 or -1. Since all terms are positive, the limit (again, if there is a limit) must be 2.

Two show that this is convergent (so that the above is correct) you can show this sequence is increasing and bounded above. Since we got 2 as the limit above, it makes sense to show that $S_n< 2$ for all n. Yes, I would do that using induction on n. When n= 1, $S_n= \sqrt{2}< 2$. Assume that, for some k, $S_k< 2$. Then $S_{k+1}= \sqrt{2+ S_k}< \sqrt{2+ 2}= \sqrt{4}= 2$.