$s_{n+1} = \sqrt{s_n+1}$ converges to $\frac{1+\sqrt{5}}{2}$

Question

Let $s_n = 1$ and for $n \geq 1$, let $s_{n+1} = \sqrt{s_n+1}$. Given this sequence and assume that it converges. I have to prove that the limit is $\frac{1+\sqrt{5}}{2}$

By using the definition of limit, I tried to set $n > N\ implies\ |\sqrt{s_n+1}-\frac{1+\sqrt{5}}{2}| < \epsilon$. But I have no idea how to simplify the expression in the absolute value and get rid of absolute value!!! Can anyone give some hints??

Call the limit $L$. Note $L>0$. Show (using the formula for $s_n$) that $L=\sqrt {L+1}$. Solve for $L$. — David Mitra, Sep 05 '13 at 15:32
I believe the first sentence should begin "Let $s_0=1$...", not $s_n$. — G Tony Jacobs, Sep 05 '13 at 15:33

André Nicolas · Accepted Answer · 2013-09-05T16:07:45.737

3

If we are forced to use an $\epsilon$-$N$ calculation, here is how it could go.

Let $\tau=(1+\sqrt{5})/2$. Then $$s_{n+1}-\tau =\sqrt{s_n+1}-\tau=\frac{s_n+1-\tau^2}{\sqrt{s_n+1}+\tau}.$$ (We multiplied "top" and "bottom" by $\sqrt{s_n+1}+\tau$.)

Now use the fact that $1-\tau^2=-\tau$, and that $\sqrt{s_n+1}+\tau \gt 1+\tau\gt 2$ to get the inequality $$|s_{n+1}-\tau| \lt \frac{1}{2}|s_n-\tau|.\tag{1}$$ Note that $|s_1-\tau|\lt 1$. Using (1) we then obtain $$|s_{k}-\tau|\lt \frac{1}{2^{k-1}}.$$ This estimate now can be used in a routine way to find a suitable $N$, given $\epsilon$.

Remark: We deliberately gave away a lot in the calculation. One can sharpen estimates to get precise information about the rate of convergence.

edited Sep 05 '13 at 16:07

answered Sep 05 '13 at 15:57

André Nicolas

507,029

Thanks for your answer..!! There are few parts that I dont understand!! (1). How is $1-\tau^2 = -\tau$?? (2). How is $|s_{n+1}-\tau < \frac{1}{2}|s_n-\tau| \rightarrow |s_k-\tau| < \frac{1}{2^{k-1}}$?? Thanks!! – eChung00 Sep 05 '13 at 16:16
First question: $1-\tau^2=-\tau$ because $\tau$ is a solution of the equation $x^2-x-1=0$ (use the Quadratic Formula). Second question: Inequality (1) ways that the "error" goes down by a factor of at least $\frac{1}{2}$ for each iteration. Since the error at $s_1$ is $\lt 1$, the error at $s_2$ is $\lt \frac{1}{2}$, the error at $s_3$ is $\lt \frac{1}{4}$, and so on. – André Nicolas Sep 05 '13 at 16:23
You are welcome. – André Nicolas Sep 06 '13 at 17:20

score 0 · Answer 2 · answered Sep 05 '13 at 15:41

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If you can assume that it is convergent, you can use continuity of the function $\sqrt{x+1}$ to conclude $$s:=\lim s_{n+1}=\lim\sqrt{s_n+1}=\sqrt{s+1}$$

answered Sep 05 '13 at 15:41

Simon Markett

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score 0 · Answer 3 · answered Sep 05 '13 at 15:42

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Hint

$$L=\lim_{n\to\infty} s_n=\lim_{n\to\infty} s_{n+1}$$

answered Sep 05 '13 at 15:42

Salech Alhasov

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$s_{n+1} = \sqrt{s_n+1}$ converges to $\frac{1+\sqrt{5}}{2}$

3 Answers3