Let $a_n= \sqrt{1+ \sqrt{2+\cdots+ \sqrt{n} } }$
How find $\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$ ?
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This expression meaningless – Michael Galuza Jul 16 '15 at 16:57
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the first to find $$\lim_{n\to\infty}\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}$$ – piteer Jul 16 '15 at 17:15
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if $lim a_n=g$ them to find $\lim_{n \to \infty } ( \sqrt{n} \cdot \sqrt[n]{g-a_n}) $ – piteer Jul 16 '15 at 17:17
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I thought that $\lim_{n\to\infty} a_n=+\infty$ – Michael Galuza Jul 16 '15 at 17:19
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@MichaelGaluza why? – piteer Jul 16 '15 at 17:26
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2@piteer: look at http://mathworld.wolfram.com/NestedRadicalConstant.html. $\lim_{n\to +\infty} a_{n}$ does not have a nice closed form expression. The answer just depends on the speed of convergenge of your sequence. – Jack D'Aurizio Jul 16 '15 at 17:34
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$a_\infty$ is the nested radical constant, whose decimal expansion can be found here. – Lucian Jul 16 '15 at 17:35
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1The limit is $\sqrt{e}/2$ .. but I need to figure out a way to write a nice proof for it ,.. – r9m Jul 16 '15 at 17:39
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@r9m lim $a_n= \sqrt{e}/2$? – piteer Jul 16 '15 at 17:44
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@piteer no no .. $\lim_{n \to \infty } ( \sqrt{n} \cdot \sqrt[n]{( \lim_{n \to \infty } a_n)-a_n}) = \sqrt{e}/2$, – r9m Jul 16 '15 at 17:45
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can you show a prove? – piteer Jul 16 '15 at 17:46
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For $a_n=\sqrt{1\sqrt{2\cdots\sqrt{n}}}$ we have $\displaystyle\lim_{n\to\infty}\sqrt[\large^n]{a_\infty-a_n}=\frac12$ – Lucian Jul 17 '15 at 00:44
1 Answers
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Call the limit $\lim\limits_{n\to \infty} a_n = a$.
Let us denote: $$a_{k,n} = \sqrt{k+\sqrt{k+1+\sqrt{\cdots + \sqrt{n}}}}$$
Since, $\displaystyle (a_{k,n+1} - a_{k,n}) = \frac{a_{k+1,n+1} - a_{k+1,n}}{a_{k,n+1}+a_{k,n}}$ we have: $ \displaystyle (a_{n+1} - a_n) = \frac{\sqrt{n+1}}{\prod\limits_{k=1}^{n}(a_{k,n+1}+a_{k,n})}$
Now, $\displaystyle a_{k,n} = \sqrt{k+a_{k+1,n}}$, where, $1 \le k \le n-1$.
Note that we have $\sqrt{k} \le a_{k,n} \le \sqrt{k} + 1$, so that plugging $a_{k,n} = \sqrt{k}+\mathcal{O}(1)$ in the recursive relation we get
$$\begin{align}a_{k,n} = \sqrt{k}\left(1+\dfrac{a_{k+1,n}}{k}\right)^{1/2} &= \sqrt{k}\left(1+\frac{\sqrt{k+1}+\mathcal{O}(1)}{k}\right)^{1/2}\\ &= \sqrt{k} + \frac{1}{2} + \mathcal{O}\left(\frac{1}{\sqrt{k}}\right)\end{align}$$
Again, $$\begin{align}a_{k,n} = \sqrt{k}\left(1+\frac{a_{k+1,n}}{k}\right)^{1/2} &= \sqrt{k}\exp \left\{\frac{1}{2}\log \left(1 + \frac{a_{k+1,n}}{k}\right)\right\} \\ &= \sqrt{k}\exp \left\{\frac{1}{2} \left(\frac{a_{k+1,n}}{k} - \frac{a_{k+1,n}^2}{2k^2} + \mathcal{O}\left(\frac{a_{k+1,n}^3}{k^3}\right)\right)\right\} \\ &= \sqrt{k}\exp\left\{\frac{1}{2\sqrt{k}} + \mathcal{O}\left(\frac{1}{k^{3/2}}\right)\right\}\end{align}$$
Hence, $$\begin{align}a_{n+1} - a_{n} &= \frac{(1+\mathcal{o}(1))\sqrt{n+1}}{2^{n}(n!)^{1/2}}\exp\left\{-\sum\limits_{k=1}^{n}\left(\dfrac{1}{2\sqrt{k}} + \mathcal{O}\left(\dfrac{1}{k^{3/2}}\right)\right)\right\}\\&= \frac{(1+\mathcal{o}(1))\sqrt{n+1}}{2^{n}(n!)^{1/2}}\exp\left\{-\sqrt{n} + \mathcal{O}(1)\right\}\end{align}$$
Therefore, $$a - a_N = \sum\limits_{n\ge N} (a_{n+1} - a_n) = e^{\mathcal{O}(1)}\sum\limits_{n \ge N}\frac{\sqrt{n+1}}{2^n(n!)^{1/2}}e^{-\sqrt{n}}$$
Now, by Stirling's Approximation we have:
$$\left(\frac{\sqrt{n+1}}{2^n(n!)^{1/2}}e^{-\sqrt{n}}\right)^{1/n} \sim \frac{\sqrt{e}}{2}e^{-\frac{1}{2}\log n}$$
Hence, $\displaystyle \lim\limits_{n \to \infty}\sqrt{n}\cdot \sqrt[n]{a-a_n} = \lim\limits_{n \to \infty} \sqrt{n} \cdot \sqrt[n]{a_{n+1}-a_n} = \frac{\sqrt{e}}{2}$
r9m
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1In the line where you get the estimate $\sqrt{k}\left(1+\dfrac{1}{2\sqrt{k}} +\mathcal{O}\left(\dfrac{1}{k}\right)\right)$, don't we need to have $a_{k+1,n}=\sqrt{k}+\mathcal{O}(1)$? Have we shown this? Also in the next line, the error drops from $\mathcal{O}\left(\frac1k\right)$ to $\mathcal{O}\left(\frac1{k^{3/2}}\right)$ that doesn't seem right. – robjohn Jul 16 '15 at 22:59
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1Where does the "exp" in the third approximation of $a_{k,n}$ come from? I obtained $\text{exp}{\frac{1}{2\sqrt{k}} + O(\frac{1}{k^{3/2}})}$.
Also, the exponent $1/n$ in $(a-a_N)^{1/n}$ just goes through the sum, this doesn't seem correct to me.
– Kolja Jan 03 '22 at 15:02 -
1@Kolja You are right .. I must have missed a term from higher order terms .. added a line for the 'exp' step. Also the reason for checking the asymptotics of $b_n^{1/n} := (a_{n+1} - a_n)^{1/n} \sim \frac{\sqrt{e}}{2\sqrt{n}}$ is that then we know that for large $n$ the ratio $\frac{b_{n+1}}{b_n}$ behaves like $b_n^{1/n}$ which decays faster than any geometric series .. hence the equality of the last limit. – r9m Jan 03 '22 at 16:05