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Does the sequence $s_{n+1}=\sqrt{1+s_n}$ always converge, no matter what the initial value of $s_1$ is?

Is this sequence always increasing and bounded? I think so, but what's throwing me off is that to find what the sequence converges to, we just solve $s^2=1+s$ to get $s=\frac{1+\sqrt{5}}{2}$.

How can the sequence converge to this number if its inital value is $s_1=3$ for example?

user162520
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2 Answers2

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For the case $s_1 = 3$, then $s_2 = 2 < 3 = s_1$ and $f(x) = \sqrt{1+x}\implies f'(x)=\dfrac{1}{2\sqrt{1+x}}> 0\implies f$ is an increasing function $\implies s_n$ is strictly decreasing sequence and is bounded below by $0$ as $s_n > 0, \forall n\ge 1$, hence is convergent to $L$ which is the solution of $L = \sqrt{1+L}\implies L = \dfrac{1+\sqrt{5}}{2}$ as claimed. In general, $s_1 \ge -1$ to begin with. Now if $\dfrac{1-\sqrt{5}}{2} < s_1 < \dfrac{1+\sqrt{5}}{2}\implies s_2 > s_1$ and the sequence is strictly increasing and is bounded above by $\dfrac{1+\sqrt{5}}{2}$ which can be shown by induction on $n \ge 1$. Thus it converges to $L =\dfrac{1+\sqrt{5}}{2}$ again. If $s_1 = \dfrac{1+\sqrt{5}}{2}$, then $s_n = \dfrac{1+\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L =\dfrac{1+\sqrt{5}}{2}$ also. If $s_1 > \dfrac{1+\sqrt{5}}{2}\implies s_n$ is a decreasing sequence as before and is bounded below by $0$ so is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ because $L \ge 0$. If $s_1 = \dfrac{1-\sqrt{5}}{2}\implies s_n = \dfrac{1-\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L = \dfrac{1-\sqrt{5}}{2}$ as $n \to \infty$. Finally, if $-1 \le s_1 < \dfrac{1-\sqrt{5}}{2}\implies s_1 >s_2\implies s_n$ is a strictly decreasing sequence, and is bounded below by $0$ hence is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ since $L \ge 0$. This completes the analysis regarding the possible values of the initial term $s_1$.

  • If $f$ is increasing then isn't $s_n$ increasing? I see now that the sequence is always monotone though. So $s_n$ is always a convergent sequence no matter what the initial value is, as long as its bigger than $-1$, and it always converges to $\frac{1+\sqrt{5}}{2}$? That's pretty cool – user162520 Jun 14 '20 at 01:03
  • @user162520: It sure is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ most of the times except when $s_1 = \dfrac{1-\sqrt{5}}{2}$. –  Jun 14 '20 at 03:44
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It is clear that for $n\ge 1$, $s_n>0$.

Let $ s$ such that $$s=\sqrt{1+s} (>0)$$ then

$$s_{n+1}-s=\sqrt{s_n+1}-\sqrt{s+1}$$

$$=\frac{s_n-s}{\sqrt{s_n+1}+\sqrt{s+1}}$$

thus

$$|s_{n+1}-s|\le \frac{|s_n-s|}{2}$$ $$\le \frac{1}{2^{n}}|s_1-s|$$

the sequence always converges to $s$.

hamam_Abdallah
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