Proving convergence of a series and then finding limit

Question

I need to show that $\sqrt{2}$, $\sqrt{2+\sqrt{2}}$, $\sqrt{2+\sqrt{2+\sqrt{2}}},\ldots$ converges and find the limit.

I started by defining the sequence by $x_1=\sqrt{2}$ and then $x_{n+1}=\sqrt{2+x_n}$. Then I proved by induction that the sequence is increasing and that it is bounded. Then I use the Monotone Convergence Theorem to prove that it converges.

Now, I claim that the limit of the sequence is $2$. So I need to show that for all $\epsilon>0$, there exists $N$ such that for all $n\geq N$ we have $|x_n-2| < \epsilon$. This is kind of where I'm stuck. I don't know how to proceed to prove the convergence to $2$.

$x_{n+1}$ converges to $x^$, $x_n$ converges to $x^$, and the square root is continuous. — user251257, Sep 02 '15 at 19:30
Since $x_{n+1}=\sqrt{2+x_n}$ and $\lim a_n=x^$, by continuity we know $\sqrt(2+x^)=x^*$. — user111463, Sep 02 '15 at 19:31

score 2 · Answer 1 · answered Sep 02 '15 at 19:30

2

Prove that this sequence is increases and bounded. Hence it convergence to some point $p\in \mathbb{R}$ and make limit transition in $x_{n+1}=\sqrt{2+x_n}$

answered Sep 02 '15 at 19:30

RFZ

16,814

score 2 · Answer 2 · answered Sep 02 '15 at 19:37

Let $f(x)=\sqrt{2+x}$, $x\geq 0$. Note that $f(x)-f(y)=\sqrt{2+x}-\sqrt{2+y}=\frac{x-y}{\sqrt{2+x}+\sqrt{2+y}}$. So $|f(x)-f(y)\leq \frac{1}{2\sqrt{2}}|x-y|$.Then $f$ in an contraction and there is $\lim_n f^n(x)$, independently of $x$. If $x=x_n$ you have $f(x)=x_{n+1}$. Given $x=x_0$ you have $x_n=f^n(x_0)$. This implies that $x_n$ is convergent.Your limit satisfy $L^2=2+L$.

score 0 · Answer 3 · 2015-09-02T20:34:08.550

0

You should know: An increasing bounded above sequence is convergent.

Now, let $R_n= \underbrace{\sqrt {2+\sqrt {2+\sqrt {\dots +\sqrt 2}}}}_{n\: \text {number of square roots}}$

Now, note that $R_1=\sqrt 2\lt2$

$R_2=\sqrt {2+\sqrt 2}\lt\sqrt {2+2}=2$

$R_3=\sqrt {2+\sqrt {2+\sqrt 2}}\lt2$ and so on.

Hence $R_n$ is an increasing and bounded above sequence {the monotonicity is obvious}.

So, let $R_n\rightarrow l$ (say)

Again $R^2_{n+1}=2+R_n$

$\Rightarrow \lim \limits_{n\rightarrow \infty} R^2_{n+1}=\lim \limits_{n\rightarrow \infty}(2+R_n) $

$\Rightarrow l^2-l-2=0$

leading to $l=-1$ or $l=2$. Since $l=-1$ is absurd,the result is $l=2$

edited Sep 02 '15 at 20:34

answered Sep 02 '15 at 19:52

You mean $l = -1$ or $l=2$... – peter a g Sep 02 '15 at 20:32
Oh, I'm sorry. I am correcting this. Thank you for this observation.@peter a g – Sep 02 '15 at 20:33

peter a g · Answer 4 · 2015-09-02T20:36:24.940

0

Somewhat in the same spirit as Euler88's answer: $$ | x_{n+1} - 2 | = | \sqrt{ 2 + x_n} - 2 |= {| 2+x_n -4 | \over \sqrt {2+ x_n } + 2 } \le {| x_n -2 | \over 2},$$ (assuming $x_n \ge 0$ - the second equality is 'rationalization in reverse'). Hence $|x_{n+1}- 2| \le |x_1- 2|/2^n$, and $x_n$ converges to $2$.

edited Sep 02 '15 at 20:36

answered Sep 02 '15 at 19:56

peter a g

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Proving convergence of a series and then finding limit

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