I recently came across this problem

Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }} = 3$

After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.

So my question is

Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.

Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .

If the solution from that book is required I can post it but it is not complete as I mentioned.

Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.

  • 2
    $\begingroup$ Did you look at all the links in all the answers to that earlier question? $\endgroup$ Jul 2, 2012 at 12:27
  • $\begingroup$ Thanks I found the answer in this article Herschfeld: On infinite radicals. $\endgroup$
    – sabertooth
    Jul 2, 2012 at 12:59
  • 1
    $\begingroup$ @GerryMyerson I think it's hard to search for the earlier posts. $\endgroup$
    – Yai0Phah
    Jul 2, 2012 at 13:18
  • $\begingroup$ @Frank While it is true that the SE search function leaves much to be desired (instead try googling with site:MSE), it does work well for reasonable unique terms like "nested radical". Indeed, it yields the cited duplicate question as first match. But, of course, one does need to know the English buzzwords for these objects. $\endgroup$ Jul 2, 2012 at 14:46
  • $\begingroup$ @Frank, yes, it's hard to search for earlier posts, and no criticism of OP is implied when someone points out that a question has been asked and answered before. But perhaps I miss the point of your comment? Anyway, now that we all know it's a duplicate, why has no one else voted to close? $\endgroup$ Jul 2, 2012 at 23:52

5 Answers 5


Note that $$ x+1=\sqrt{1+x(x+2)}\tag{1} $$ Iterating $(1)$, we get $$ x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\color{#C00000}{(x+5)}}}}}\tag{2} $$ Note that $$ s_3=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\color{#C00000}{\sqrt{1}}}}}}\tag{3} $$ Instead of $\color{#C00000}{\sqrt{1}}$ as in the last term of $(3)$, $(2)$ has $\color{#C00000}{(x+n+2)}$. Thus, the increasing sequence in $(3)$ is bounded above by $x+1$. Thus, the sequence in $(3)$ has a limit.

  • 1
    $\begingroup$ I try to compliment this every time - but I love the custom-dulled colors. $\endgroup$
    – davidlowryduda
    Dec 15, 2012 at 19:40
  • $\begingroup$ @mixedmath Yes, they feel quite soft and seems fitting for math work. $\endgroup$ Dec 6, 2016 at 22:15

Let $f_n(0)=\sqrt{1+n}$ and $f_n(k)=\sqrt{1+(n-k)f_n(k-1)}$. Then $0<f_n(0)<n+1$ when $n>0$. Assume that $f_n(k)<n+1-k$ and we can show by induction that $$ f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1) $$ for all k. Your expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$, so converges.


$x + 1 = \sqrt {1 + x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{}.....}}}$. Put $x=2$ gives you the solution. For proof see http://zariski.files.wordpress.com/2010/05/sr_nroots.pdf

  • $\begingroup$ It is the same proof that I know but discussion of convergence is not complete still +1 for dicussion of numerical convergence. $\endgroup$
    – sabertooth
    Jul 2, 2012 at 12:52

Note that $\frac{x+y}{x+z} < \frac{y}{z}$ for positive $x, y, z$; thus $\frac{\sqrt{x+y}}{\sqrt{x+z}} < \frac{\sqrt{y}}{\sqrt{z}}$ for $y > z$.

Thus we get $\frac{a_{n+1}}{a_n} = \frac{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1+(n+1)\sqrt{1}}}}}}}{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}} < \frac{\sqrt{\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1+(n+1)\sqrt{1}}}}}}}{\sqrt{\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}} < \frac{\sqrt{\sqrt{\cdots\sqrt{n+2}}}}{\sqrt{\sqrt{\cdots\sqrt{1}}}} = O(n^{(2^{-n})})$.

Next, $\ln{a_{n+1}}-\ln{a_n} = O(\frac{\ln_n}{2^n})$. Summing the equations we get $\ln{a_n} - \ln{a_1} = O(\frac{\ln_1}{2^1} + \cdots + \frac{\ln_n}{2^n})$; letting $n$ to approach $\infty$, we get $\ln{a} - \ln{a_1} = O(1)$, thus there is a finite limit of $a_i$.


This is a known question of Putnam competition: I give a complete proof with details:


it is clear that for $m>n$, we have $f_m(x)>f_n(x)$. Moreover

$x+1=\sqrt{1+x(x+2)}=\sqrt{1+x\sqrt{1+(x+1)(x+2)}}=\cdots=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\cdots+(x+n-1)(x+n+1)}}}\geq f_n(x)$.

So this tells us that $\lim_{n\to \infty}f_n(x)$ exists. Now let $\lim_{n\to \infty}f_n(x)=f(x)$ hence $f(x)\leq x+1$ and $f(x)>\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+...}}}}=\frac{x+\sqrt{x^2+4}}{2}>x$ hence $x<f(x)<x+1$. Now, let $g(x)=x+1-f(x)$ then $0\leq g(x)<1$, it is easy to see that $(f(x))^2=1+xf(x+1)$ hence $(x+1+f(x))g(x)=xg(x+1)$, so

$$\frac{g(x)}{x}\leq \frac{g(x+1)}{x+1}\leq \cdots\leq \frac{g(x+n)}{x+n}<\frac{1}{x+n}\to 0$$ hence $f(x)=x+1$, so $f(2)=3$


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