A sequence of nested radicals and its limit

Question

Please Help me in the following Problem

What is the Number Of Natural Numbers ,$n\le30$ for which $\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}$ is also a prime number.

The only way I am able to find to solve this is calculate each and every term once but it will be extremely lengthy. Please suggest me a shorter way

NOTE This is a NSEJS (STAGE-1) Problem

Suppose that $\sqrt{n+\sqrt{n+\sqrt{n+\dots}}}$ actually converges to a real number $S$ (This assumption should still be proven to have a complete argument). Then $S^2=\left(\sqrt{n+\sqrt{n+\sqrt{n+\dots}}}\right)^2=n+S$, and rearranged that $S^2-S-n=0$. Can you continue? — JMoravitz, Jan 24 '18 at 06:43
@JMoravitz I have reached the following conclusion Let $\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}=p$ where $p$ is a prime number $p^2=n+p$ so $p=\sqrt{n+p}$ Now putting $p$ a prime one by one since there are only 11 such prime numbers we will get all the values of n for which we get the results.Am I Right? — S.Bansal, Jan 24 '18 at 06:49
Rewriting again, we have $n=S(S-1)$. How many values of $n\geq 30$ can be written as a product of a prime and a number one less than that prime? (How many prime numbers are there? Is there a value of $n$ corresponding to each?) — JMoravitz, Jan 24 '18 at 06:51
My final answer is that there are exactly 3 such prime numbers . This way it is extremely small as after the third integer all are coming larger than 30 — S.Bansal, Jan 24 '18 at 06:57
Perhaps you have forgotten to list an important condition or you have an inequality pointing in the wrong direction or with the wrong expression in the inequality. Are you for some reason looking for $n\color{red}{\leq}30$ instead? — JMoravitz, Jan 24 '18 at 07:00
Could you please explain why so that i could correct my fault @miracle173 — S.Bansal, Jan 24 '18 at 07:15
do you want to know the number of natural numbers n<= 30 where this square root expression is a prime number? — miracle173, Jan 24 '18 at 07:20

fred goodman · Accepted Answer · 2018-01-24T20:52:33.870

If $n = p^2 -p$, then $p^2 =n + p $, so $p = \sqrt{n + p}= \sqrt{n + \sqrt{n+p}}$, etc.

Edit: My original answer, above, left a lot to the imagination. The real work would be in analyzing the convergence of the sequence of nested radicals. To this end, fix a positive number $n$, not necessarily an integer for now. Define $a_k = a_k(n)$ by $a_1 = \sqrt{n}$, and $a_{k+1} = \sqrt{n + a_k}$ for $k \ge 1$.

Claim:

For fixed $k$, $a_k(n)$ is an increasing function of $n$.
For fixed $n$, $(a_k(n))_{k \ge 1}$ is an increasing sequence and bounded above, hence convergent to a limit $p = p(n)$.

The first claim can be proved by induction on $k$. For the second claim, note that $a_2 - a_1 > 0$, and $$ (a_{k+1} - a_k)(a_{k+1} + a_k) = a_k - a_{k-1} $$ for $k >1$. By induction on $k$, $a_{k+1} - a_k > 0$ for all $k$. Now assume for the moment that $n \ge 1$. Then $a_k \ge a_1 = \sqrt{n} \ge 1$. Hence $$ (a_{k+1} - a_k) \le (1/2)( a_k - a_{k-1}) $$ for $k >1$. It follow that the sequence $(a_k(n))_{k \ge 1}$ is convergent. But now if $0 < n < 1$, $a_k(n) \le a_k(1) \le \lim_{k\to \infty} a_k(1)$, so again the sequence $(a_k(n))_{k \ge 1}$ is bounded above, and hence convergent. Let $p = p(n)$ denote $\lim_{k\to \infty} a_k(n)$. Again, $p$ is not necessarily an integer let alone a prime. This completes the proof of the claims.

Now fix $n>0$, and let $p = \lim_{k\to \infty} a_k(n)$. Then we have $$ p = \sqrt{n + p}, $$ as follows from $a_{k+1} = \sqrt{n + a_k}$. Therefore $p^2 - p = n$, and thus $p$ is the unique positive root of the quadratic equation $p^2 - p = n$. Conversely, if we start with a positive number $p > 1$ and put $n = p^2 - p$, then the sequence $a_k(n)$ converges to $p$. Thus $p$ is the limit of the sequence of nested radicals $a_k(n)$ if and only if $p^2 - p = n$.

Now we can solve the original question by searching for primes $p$ such that $p^2 - p \le 30$.

score 1 · Answer 2 · 2018-01-24T07:39:23.577

1

Let the limit be $x$. Then $\sqrt{n+x}=x$. Thus $x^2-x-n=0$... So $x=\frac{1\pm\sqrt{1+4n}}2$. $x$ is positive, so $x=\frac{1+\sqrt{1+4n}}2$.

So let's see: $n=6 \implies x=3$

Say $x$ is a prime, $p$. Then $p^2-p-n=0$, so $n=p(p-1)$...

It looks like infinitely many... (as there are infinitely many primes)

edited Jan 24 '18 at 07:39

answered Jan 24 '18 at 07:07

I may be wrong but I can only count three primes, $p = 2, 3, 5$. – user8277998 Jan 24 '18 at 11:22

A sequence of nested radicals and its limit

2 Answers2