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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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  • $\begingroup$ This expression meaningless $\endgroup$ – Michael Galuza Jul 16 '15 at 16:57
  • $\begingroup$ the first to find $$\lim_{n\to\infty}\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}$$ $\endgroup$ – piteer Jul 16 '15 at 17:15
  • $\begingroup$ if $lim a_n=g$ them to find $\lim_{n \to \infty } ( \sqrt{n} \cdot \sqrt[n]{g-a_n}) $ $\endgroup$ – piteer Jul 16 '15 at 17:17
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    $\begingroup$ @piteer: look at 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. $\endgroup$ – Jack D'Aurizio Jul 16 '15 at 17:34
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    $\begingroup$ The limit is $\sqrt{e}/2$ .. but I need to figure out a way to write a nice proof for it ,.. $\endgroup$ – r9m Jul 16 '15 at 17:39
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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, $1 \le k \le n-1$.

We have first: $\sqrt{k} \le a_{k,n} \le \sqrt{k} + 1$, so that $a_{k,n} = \sqrt{k}+\mathcal{O}(1)$

Hence, $$\begin{align}a_{k,n} = \sqrt{k}\left(1+\dfrac{a_{k+1,n}}{k}\right)^{1/2} &= \sqrt{k}\left(1+\frac{\sqrt{k+1}+\mathcal{O}(1)}{k}\right)^{1/2}\\ &= \sqrt{k}\left(1+\dfrac{1}{2\sqrt{k}}+\mathcal{O}\left(\dfrac{1}{k}\right)\right)\end{align}$$

Again, $$\begin{align}a_{k,n} = \sqrt{k}\left(1+\frac{a_{k+1,n}}{k}\right)^{1/2} &= \sqrt{k}\exp\left\{\frac{1}{2k}\left(\sqrt{k}+\frac{1}{2}+\mathcal{O}\left(\frac{1}{\sqrt{k}}\right)\right)\right\}\\&= \sqrt{k}\exp\left\{\frac{1}{2\sqrt{k}}+\frac{1}{2k}+\mathcal{O}\left(\frac{1}{k^{3/2}}\right)\right\}\end{align}$$

Hence, $$\begin{align}a_{n+1} - a_{n} &= \frac{(1+\mathcal{o}(1))\sqrt{n+1}}{2^{n}(n!)^{1/2}}\exp\left\{-\sum\limits_{k=1}^{n}\left(\dfrac{1}{2\sqrt{k}}+\frac{1}{2k}+\mathcal{O}\left(\dfrac{1}{k^{3/2}}\right)\right)\right\}\\&= \frac{(1+\mathcal{o}(1))\sqrt{n+1}}{2^{n}(n!)^{1/2}}\exp\left\{-\sqrt{n} - \frac{1}{2}\log n+\mathcal{O}(1)\right\}\end{align}$$

Therefore, $$a - a_N = \sum\limits_{n\ge N} (a_{n+1} - a_n) = e^{\mathcal{O}(1)}\sum\limits_{n \ge N}\frac{\sqrt{n+1}}{2^n(n!)^{1/2}}e^{-\sqrt{n}-\frac{1}{2}\log n}$$

Now, by Stirling's Approximation we have:

$$\left(\frac{\sqrt{n+1}}{2^n(n!)^{1/2}}e^{-\sqrt{n}-\frac{1}{2}\log n}\right)^{1/n} \sim \frac{\sqrt{e}}{2}e^{-\frac{1}{2}\log n}$$

Hence, $\displaystyle \lim\limits_{n \to \infty}\sqrt{n}\cdot \sqrt[n]{a-a_n} = \lim\limits_{n \to \infty} \sqrt{n} \cdot \sqrt[n]{a_{n+1}-a_n} = \frac{\sqrt{e}}{2}$

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    $\begingroup$ In 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. $\endgroup$ – robjohn Jul 16 '15 at 22:59
  • $\begingroup$ @robjohn fixed. That part at most adds a contribution of $\frac{1}{2}\log n$, but when we take Stirling's approximation it is dominated by $\sqrt{n}$, so the final result remains same. Thanks! $\endgroup$ – r9m Jul 17 '15 at 4:19

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