This is to prove how the limit of $s_n$ converges to $\frac{1}{2}(1+\sqrt{5})$.
Assume: $s_1 = 1$; for $n \geq 1$, $s_{n+1} = \sqrt{s_n + 1}$.
How to prove this converges to $\frac{1}{2}(1+\sqrt{5})$?
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1Take a look at this post with answer. It is possible that you could use some or all of that to aid in your proof here. What do you think about that? Is it useful? – J. W. Perry Sep 25 '13 at 04:58
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I think you can make use of $\lim_{n \rightarrow \infty}s_{n} = \lim_{n \rightarrow \infty}s_{n+1} = a$. – Petch Puttichai Sep 25 '13 at 05:00
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@PetchPuttichai Yes, but only once we know that the limit exists. – Andrés E. Caicedo Sep 25 '13 at 05:01
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2Note that $\sqrt{s_n+1}$ is not $\frac12(1+\sqrt5)$ (for no $n$) and that a limit does not "converge", a sequence does. – Did Sep 25 '13 at 05:03
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Argue by induction:
- The terms of the sequence are all positive.
- They are all bounded above (by $3$): The base case is clear, and if $s_n<3$, then $s_{n+1}=\sqrt{s_n+1}<\sqrt{3+1}=2<3$.
- The sequence is increasing: $s_2=\sqrt2>1=s_1$, and if $s_n<s_{n+1}$, then $s_n+1<s_{n+1}+1$, so $s_{n+1}=\sqrt{s_n+1}<\sqrt{s_{n+1}+1}=s_{n+2}$.
It follows that the sequence converges (to its supremum). Call its limit $L$, so $L=\lim_n s_{n+1}=\lim_n\sqrt{s_n+1}=\sqrt{L+1}$ (by continuity of $f(x)=\sqrt{x+1}$).
Solving the equation $L=\sqrt{L+1}$ gives us that $L^2-L-1=0$, so $L=\frac12(1\pm\sqrt5)$, and the sign must be $+$ rather than $-$ since the terms of the sequence are all positive, so also their limit $L$ is non-negative, $L\ge0$.
Andrés E. Caicedo
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Consider applying the Contraction Mapping Theorem. Just check that the conditions for the theorem are satisfied (I'll leave those details to you).
We have $s_{n+1} = \sqrt{s_n + 1}$. Take $F: x \to \sqrt{x +1}$. By the Contraction Mapping Theorem we can conclude that $F$ has a unique fixed point $s$.
This gives us: \begin{align*} \lim s_{n+1} =& \lim\sqrt{s_n +1} \\ s =& \sqrt{s+1} \\ s^2 - s - 1 =& 0 \end{align*}
Then by the quadratic formula we have that our unique fixed point $s = \dfrac{1+\sqrt5}{2}$.
CodeKingPlusPlus
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