# Recursive square root inside square root problem

I have been debating this issue for days:

I can't find a recursive function of this equation:

$\large{\sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}}}$

has been trying to find a solution this for days now, is what I have achieved so far:

$f(n)=\sqrt{2 f(n-1)}, f(1)=\sqrt{2}$

Unfortunately, I do not know how to move forward, thanks a lot!

## 1 Answer

I would write it as $$f\left(n\right)=\sqrt{n+1+\pi f\left(n+1\right)},$$ Hence $$f\left(1\right)=\sqrt{2+\pi f\left(2\right)}, \ f\left(2\right)=\sqrt{3+\pi f\left(3\right)} \Rightarrow f\left(1\right)=\sqrt{2+\pi\sqrt{3+f\left(3\right)}}$$ which means you search for $f\left(1\right)$. $$f\left(n+1\right)=\frac{1}{\pi}\left(f\left(n\right)^2-n-1\right)$$

• I dont think so, wolfram interprets it as f(0)=0 – Atmos Sep 3 '18 at 9:51
• I don t think so. However do you want to find a value to this ? – Atmos Sep 3 '18 at 15:21
• There is no value for n=0 the sequence i wrote us for n>=1 – Atmos Sep 3 '18 at 15:30
• Hi, thank you for your reply. It is not correct, both in Matlab and WolframAlpha the sequence is not correct. link If someone can understand how to correct your formula would be perfect, yesterday I tried but I failed – Gionata Donati Sep 4 '18 at 8:38