Let $y_1= \sqrt{p}$ where $p > 0$, and $y_{n+1}=\sqrt{p+y_n}$ for $n \in \mathbb{N}$. Show that $(y_n)$ converges and find the limit. [Hint: One upper bound is $1+2\sqrt{p}$]
Proof:
1) The sequence is bounded:
Clearly, $y_n>0 ; \forall n \in \mathbb{N}$. This implies that $0$ is a lower bound. The upper bound of $1+2\sqrt{p}$ along with the lower bound imply that the sequence is bounded.
2) Claim: the sequence is increasing:
$y_1 = \sqrt{p}<\sqrt{p+\sqrt{p}}=y_2$
Assume $y_n<y_{n+1}$
$\implies p+y_n<p+y_{n+1}$
$\implies \sqrt{p+y_n}<\sqrt{p+y_{n+1}}$
$\implies y_{n+1}<y_{n+2}$
Thus, by PMI, $y_{n+1}>y_n$ for all $n \in \mathbb{N}$
Since the sequence is bounded and it is increasing(thus monotone), then, by the Monotone Convergence Theorem, $\lim{y_n}$ exists.
Suppose $\lim{y_n}=y$
Then $\lim{y_{n+1}}=\lim{\sqrt{p+y_n}}$
$\implies y=\sqrt{p+y}$
$\implies y^2 -y-p=0$
I don't know how to go about this anymore. How do I find the limit considering I don't know the value of p? Can anyone please explain as well as verify the work done till now?
Many thanks.
Taking...from where you stopped...we have: $y^2 - y - p = 0 \implies y = \dfrac{1 \pm \sqrt{1+4p}}{2}$. Since $p_n > 0, \forall n \ge 1 \implies L \ge 0 \implies L = \dfrac{1+\sqrt{1+4p}}{2}$, whereas $L$ is the limit of the sequence.