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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.

Martin Sleziak
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sabertooth
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    Did you look at all the links in all the answers to that earlier question? – Gerry Myerson Jul 02 '12 at 12:27
  • Thanks I found the answer in this article Herschfeld: On infinite radicals. – sabertooth Jul 02 '12 at 12:59
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    @GerryMyerson I think it's hard to search for the earlier posts. – Yai0Phah Jul 02 '12 at 13:18
  • @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. – Bill Dubuque Jul 02 '12 at 14:46
  • @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? – Gerry Myerson Jul 02 '12 at 23:52
  • @BillDubuque Well, so I think one way is to ask in the chat room first. PS: I just asked this problem in chat room, and anon pointed out that it's a problem of Ramanujan. – Yai0Phah Jul 03 '12 at 04:44
  • @Frank, it is also identified as Ramanujan's at the link I gave. – Gerry Myerson Jul 04 '12 at 10:17

5 Answers5

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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.

robjohn
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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.

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

Aang
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  • It is the same proof that I know but discussion of convergence is not complete still +1 for dicussion of numerical convergence. – sabertooth Jul 02 '12 at 12:52
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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 $a_i$.

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

$f_n(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+n-1)\sqrt{1+(x+n)}}}}$

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 $f(2)=3$