¿Cómo encontrar $\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$?

Question

Que $a_n= \sqrt{1+ \sqrt{2+\cdots+ \sqrt{n} } }$

¿Cómo encontrar $\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$?

Answer 1

Llame al límite de $\lim\limits_{n\to \infty} a_n = a$.

Nos deja denotar: $$a_{k,n} = \sqrt{k+\sqrt{k+1+\sqrt{\cdots + \sqrt{n}}}}$$

Ya, $\displaystyle (a_{k,n+1} - a_{k,n}) = \frac{a_{k+1,n+1} - a_{k+1,n}}{a_{k,n+1}+a_{k,n}}$ hemos: $\displaystyle (a_{n+1} - a_n) = \frac{\sqrt{n+1}}{\prod\limits_{k=1}^{n} a_{k,n+1}+a_{k,n})}$

Ahora, $\displaystyle a_{k,n} = \sqrt{k+a_{k+1,n}}$, donde, $1 \le k \le n-1$.

Tenemos primero: $\sqrt{k} \le a_{k,n} \le \sqrt{k} + 1$, por lo que el $a_{k,n} = \sqrt{k}+\mathcal{O}(1)$

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

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

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

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

Ahora, por Stirling Aproximación tenemos:

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

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