# Proof Verification/Help: Convergence of a recursively defined sequence

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.

• You might be expected to prove that the given upper bound is in fact an upper bound. Mar 23 '18 at 19:41
• And once you have that quadratic, you can just run the quadratic formula, yes? Mar 23 '18 at 19:42
• You can solve the quadratic equation in the last equation you get. Simply find the roots (two or one or complex roots) depending on the value of $p$. Your limit is the real value between $0$ and $1+2\sqrt p$ Mar 23 '18 at 19:45
• The proof seems correct. Tell us which parts confuse you Mar 23 '18 at 19:50
• The upper bond proof is easy by recurrence similar to the monotonicity proof Mar 23 '18 at 19:55

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.