# Finding the explicit formula given recursion

This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$

I have calculated the first 4 terms to be $$\frac12, \frac {\sqrt3}2, \frac {\sqrt{\sqrt3+2}}2, \frac {\sqrt{\sqrt{\sqrt3+2}+2}}2$$

How can I find the explicit formula?

Looking at the first 4 terms, I notice that the denominator remains at 2, while the numerator appears to also be recursive. I will call this numerator $$b_n$$ $$\begin{cases} b_1=1,\\ b_n= \sqrt {b_{n-1}+2} \end{cases}$$

I am thinking that if I find the explicit formula for $$b_n$$, perhaps I could use the formula $$\frac {b_n}2=a_n$$ and solve.

I also figure out that in the recursive formula where $$c_1=x$$, and $$c_{n}=\sqrt{c_{n-1}}$$, the explicit formula would be: $$c_n=x^\frac1{2^{(n-1)}}$$

I am currently struggling on how to solve for the case $$b_n=\sqrt{b_{n-1}+2}$$

• Can you elaborate more please? – Arjun Aug 11 '20 at 20:03
• Have you tried to compute the first values ? What does that give ? – EDX Aug 11 '20 at 20:40
• If you want to sqare the sequence, it would go like this: $b_n = a_n^2 = \frac{a_{n-1} + 1}{2} = \frac{\sqrt{b_{n-1}} + 1}{2}$. It's not enough to square the recurrence formula. That's why your graph doesn't line up. – Todor Markov Aug 11 '20 at 21:33

You can go for a limit, though: If the limit is $$A$$, it must be that:
\begin{align*} A &= \sqrt{\frac{A + 1}{2}} \\ A^2 &= \frac{A + 1}{2} \\ A &= \frac{1 \pm \sqrt{1 + 4 \cdot 2}}{4} \\ &= 1 \text{ or } -\frac{1}{2} \end{align*}
The negative value makes no sense, so the limit (if it exists) is 1. To prove the limit exists, see that $$a_n$$ is strictly increasing, limited from above by 1.