Prove that the limit of this sequence is $\frac{1}{2}(1 + \sqrt{5})$

Question

I have the following exercise: Let $S_1$ = 1 and for $n \geq 1$ let $S_{n+1} = \sqrt{S_n + 1}$. Prove that the limit of this sequence is $\frac{1}{2}(1 + \sqrt{5})$

Answer 1

To be rigorous: first check that $S_2>S_1$ and $S_1<S_0\equiv\frac{1}{2}(1+\sqrt{5})$ and use an induction argument to show that $S_n$ is increasing and each $S_n$ is bounded above by $S_0$. Thus, a limit $S$ exists satisfying $S=\sqrt{S+1}$. You can then check that $S=S_0$.

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More details: to check $S_1<S_2$, you simply plug in the numbers. Similarly, do the same for $S_1<S_0$. Now, in one induction argument, you can verify both that $S_n$ is increasing and $S_n$ is bounded above by $S_0$. We have just done the base cases. The induction steps are $$ S_{n+1}-S_n=\sqrt{S_n+1}-\sqrt{S_{n-1}+1}\geq 0\text{ if }S_n\geq S_{n-1};\\ S_{n+1}=\sqrt{S_n+1}\leq\sqrt{S_0+1}=S_0\text{ if }S_n\leq S_0. $$ Therefore, now you have $\{S_n\}$ is an increasing sequence that is bounded above. So it has some limit $S$. This $S$ satisfies $S=\sqrt{S+1}$ which you obtain by taking the limits on both sides of $S_{n+1}=\sqrt{S_n+1}$. Note that this taking limits step isn't proper unless you first justify that the limit exists. Hence, the induction argument we did in the beginning.

Answer 2

$$ S_{n+1}^2 = S_n + 1 $$ if a limit does exist then it occurs when $$ \lim_{n\rightarrow \infty}\dfrac{S_{n+1}}{S_n}=1 $$ i.e. $$ M^2 = M + 1 $$

Answer 3

The limit $L$ must satisfy $$L=\sqrt{L+1}$$

Answer 4

Hint: prove that $S_n$ is bounded and increasing after some $n$. Then, get $S_{n+1} = \sqrt{S_n + 1}$, and pass the limit to obtain $L = \sqrt{L+1}$. Then solve for $L$ (and discard the negative value you will find).