Given $A>0$, let $b_1=1$ and $b_{n+1} = \sqrt{A+b_n}$

Question

Given that $(b_n)$ converges, find its limit

The hint we received in class was to use a $b_{n+1} - b_n$ idea and then multiply by the conjugate... I have determined this is bounded, as all convergent sequences are bounded. And I believe it to be increasing just by stepping through a few iterations with different values of A

score 0 · Accepted Answer · answered Sep 29 '16 at 04:56

0

If you already know that the limit exists, then it follows \begin{align} b=\lim_{n\rightarrow \infty} b_{n+1} = \lim_{n\rightarrow \infty}\sqrt{A+b_n}= \sqrt{A+b}. \end{align} Hence we have that \begin{align} b^2-b-A = 0 \ \ \Rightarrow \ \ b = \frac{1\pm \sqrt{1+4A}}{2}. \end{align} It's not hard to check $b_n\leq b_{n+1}$. Since $b_1 = 1$, then it follows \begin{align} b= \frac{1+\sqrt{1+4A}}{2}. \end{align}

answered Sep 29 '16 at 04:56

Jacky Chong

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I'm sorry but I don't know what happened between your steps 1 and 2..... As in, where did you get $b^2-b-A=0$? – NMBL Sep 29 '16 at 05:21
Since $b = \sqrt{A+b}$, then I squared both sides. – Jacky Chong Sep 29 '16 at 05:22
Oh wow okay gotcha – NMBL Sep 29 '16 at 05:24

score 0 · Answer 2 · answered Sep 29 '16 at 05:25

The proof goes as follows:

$1$. Show that $b_n > 0$ for all $n$ using induction.

$2$. Show that $b_n \leq \dfrac{1+\sqrt{1+4A}}2$ using induction.

$3$. Hence, show that $b_n < b_{n+1}$ using induction.

$4$. Hence, by monotone sequence theorem, we know that the limit of $b_n$ exists.

$5$. Hence, $$b=\lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \sqrt{A+b_n} = \sqrt{A+\lim_{n \to \infty} b_n} = \sqrt{A+b}$$

Given $A>0$, let $b_1=1$ and $b_{n+1} = \sqrt{A+b_n}$

2 Answers2