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 $1+2\sqrt{p}$]


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.

  • $\begingroup$ You might be expected to prove that the given upper bound is in fact an upper bound. $\endgroup$
    – Valborg
    Mar 23 '18 at 19:41
  • $\begingroup$ And once you have that quadratic, you can just run the quadratic formula, yes? $\endgroup$
    – Valborg
    Mar 23 '18 at 19:42
  • $\begingroup$ 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 $1+2\sqrt p$ $\endgroup$
    – Frostic
    Mar 23 '18 at 19:45
  • $\begingroup$ The proof seems correct. Tell us which parts confuse you $\endgroup$
    – Frostic
    Mar 23 '18 at 19:50
  • $\begingroup$ The upper bond proof is easy by recurrence similar to the monotonicity proof $\endgroup$
    – Frostic
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.