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If $$a_n=\sqrt{4+\sqrt{4+3\sqrt{4+5\sqrt{4+7\sqrt{\cdots\sqrt{4+(2n-1)\sqrt{4+(2n+1)}}}}}}}$$for any natural number $n$, then evaluate $\lim_{n\to \infty}a_n$.

Note that $a_1=\sqrt{4+\sqrt{4+3}}$ and $a_2=\sqrt{4+\sqrt{4+3\sqrt{4+5}}}$.

I don't have any good idea. I need your help.

mathlove
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1 Answers1

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Define $$ f(x)=\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+(x+4)\sqrt{4+\dots}}}} $$ then $f(x)^2=4+xf(x+2)$. This indicates we should look at $f(x)=x+2$.

Considering $f(x)=x+2$, we are lead to show inductively that $$ x+2=\small\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)(x+2k+2)}}}}} $$

Define $$ f_{k,x}(y)=\left\{\begin{array}{} y&\text{if }k=0\\ \sqrt{4+xf_{k-1,x+2}(y)}&\text{if }k>0 \end{array}\right. $$ unrolled, that is $$ f_{k,x}(y)=\small \sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)y}}}}} $$ Note that $f_{0,x}(x+2)=x+2$. Suppose that, for some $k\ge0$, $f_{k,x}(x+2k+2)=x+2$. Then $$ \begin{align} f_{k+1,x}(x+2k+4) &=\sqrt{4+xf_{k,x+2}(x+2k+4)}\\ &=\sqrt{4+x(x+4)}\\ &=x+2 \end{align} $$ Therefore, for all $k\ge0$, we have $f_{k,x}(x+2k+2)=x+2$.

We want to show that $$ \lim_{k\to\infty}f_{k,x}(1)=x+2 $$ Note that $$ \frac{f_{0,x+2k}(x+2k+2)}{f_{0,x+2k}(1)}=\frac{x+2k+2}{1} $$ Suppose that, for some $j\ge0$, $$ \frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}\le(x+2k+2)^{1/2^j} $$ Then $$ \begin{align} \frac{f_{j+1,x+2k-2j-2}(x+2k+2)}{f_{j+1,x+2k-2j-2}(1)} &=\sqrt{\frac{4+(x+2k-2j-2)f_{j,x+2k-2j}(x+2k+2)} {4+(x+2k-2j-2)f_{j,x+2k-2j}(1)}}\\ &\le\sqrt{\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}}\\ &\le(x+2k+2)^{1/2^{j+1}} \end{align} $$ Therefore, for all $j\ge0$, $$ \frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}\le(x+2k+2)^{1/2^j} $$ in particular, for $j=k$, $$ \frac{f_{k,x}(x+2k+2)}{f_{k,x}(1)}\le(x+2k+2)^{1/2^k} $$

Since $f_{k,x}(x+2k+2)=x+2$, we have $$ (x+2)(x+2k+2)^{-1/2^k}\le f_{k,x}(1)\le (x+2) $$ and by the squeeze theorem, we have $$ \lim_{k\to\infty}f_{k,x}(1)=x+2 $$ as desired. Setting $x=1$, we get the answer to the question to be $3$.

robjohn
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  • I was about to write this very same answer when I realized I had no idea how to get from $f(x)^2=4+xf(x+2)$ to $f(x) = x+2$. Care to elaborate? How would you argue that $f$ has to be linear? – Arthur Aug 05 '13 at 08:39
  • Why the $\implies$? – Did Aug 05 '13 at 08:39
  • +1 for your second part, and still I got no idea to get from functional equation to explicit formula even if we assume $f$ is analytic. – asatzhh Aug 05 '13 at 08:59
  • I'm still not completely statisfied. How do we know that the two limits $$ 3 = \lim_{k\to \infty}\sqrt{4+1\sqrt{4+3\sqrt{4+\dots+\sqrt{4+(2k-1)\sqrt{4+{(2k+1)(2k+5)}}}}}} $$ and $$ ? = \lim_{n\to \infty}\sqrt{4+1\sqrt{4+3\sqrt{4+5\sqrt{4+7\sqrt{\cdots\sqrt{4+(2n-1)\sqrt{4+(2n+1)}}}}}}} $$ are the same? – Arthur Aug 05 '13 at 09:23
  • @Arthur: I have written that up now. – robjohn Aug 05 '13 at 09:25
  • @robjohn: Thank you very much for great proof! – mathlove Aug 08 '13 at 07:18
  • Thanks robjohn. I know I can count on you :-) – Venus Nov 19 '14 at 18:47