$f(n)=\sqrt{m+f(n-1)}, f(1)=\sqrt{m}.$ Find the general term of $f(n).$
Linked question: How can we prove $\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}}=2\cos\left(\frac{\pi}{2^{n+1}}\right)$ without induction
I just answered the question above, and another question popped into my head. I was curious about the general term of $\sqrt{m+\sqrt{m+\cdots\sqrt{m+\sqrt{m}}}}$, instead of $2$. So, I tried some:
\begin{align} &\displaystyle \lim_{n \to \infty} f(n)^2=m+\lim_{n\to\infty}f(n). \\ \therefore \; & \lim_{n \to \infty} f(n)=\dfrac{1 \pm \sqrt{4m+1}}{2}. \\ & f(n)>0. \Rightarrow \lim_{n \to \infty}f(n)=\dfrac{1+\sqrt{4m+1}}{2}. \\ \ \\ &f(n) \text{ increases.} \\ \therefore \; & \sqrt{m} \leq f(i) \leq \dfrac{1+\sqrt{4m+1}}{2} \text{ for } \forall i. \end{align}
How can we proceed?