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Define the sequence $\{a_n\}$as follows \begin{align*} a_n=\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} \end{align*} Apparently we can prove $\lim_{n\to \infty}a_n$exists(see How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists? we denote it by $l$. Then prove that \begin{align*} \lim_{n\to \infty}\sqrt{n}\cdot \sqrt[n]{l-a_n}=\frac{\sqrt{e}}{2}\tag{*} \end{align*}

$1^\circ$It's easy to see that$\{a_n\}$is strict increasing。In order to prove that$\lim_{n\to \infty}a_n$exists,we only need to show that $\sum_{n=1}^\infty (a_{n+1}-a_n)$converges. \begin{align*} a_{n+1}-a_n=&\sqrt{1+\sqrt{2+\cdots+\sqrt{n+\sqrt{n+1}}}}-\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}\\ =&\frac{\sqrt{2+\cdots+\sqrt{n+\sqrt{n+1}}}-\sqrt{2+\cdots+\sqrt{n}}}{\sqrt{1+\sqrt{2+\cdots+\sqrt{n+\sqrt{n+1}}}}+\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}} \end{align*} By evaluating the last term above,we get \begin{align*} a_{n+1}-a_n<\frac{1}{2\sqrt{1}}\left(\sqrt{2+\cdots+\sqrt{n+\sqrt{n+1}}}-\sqrt{2+\cdots+\sqrt{n}}\right) \end{align*} Repeating the above steps,we get \begin{align*} a_{n+1}-a_n<&\frac{1}{2\sqrt{1}}\frac{\sqrt{3+\cdots+\sqrt{n+\sqrt{n+1}}}-\sqrt{3+\cdots+\sqrt{n}}}{\sqrt{2+\cdots+\sqrt{n+\sqrt{n+1}}}+\sqrt{2+\cdots+\sqrt{n}}}\\ <&\frac{1}{2\sqrt{1}}\cdot \frac{1}{2\sqrt{2}}\left(\sqrt{3+\cdots+\sqrt{n+\sqrt{n+1}}}-\sqrt{3+\cdots+\sqrt{n}}\right)\\ <&\cdots\cdots\\ <&\frac{1}{2\sqrt{1}}\cdot \frac{1}{2\sqrt{2}}\cdots \frac{1}{2\sqrt{n}}\sqrt{n+1}=\frac{\sqrt{n+1}}{2^n\sqrt{n!}}\triangleq b_n\tag{1} \end{align*} Notice that \begin{align*} \lim_{n\to \infty}\frac{b_{n+1}}{b_n}=\lim_{n\to \infty}\frac{\sqrt{n+2}}{2^{n+1}\sqrt{(n+1)!}}\cdot\frac{2^n\sqrt{n!}}{\sqrt{n+1}}=0<1 \end{align*} Thus$\sum_{n=1}^\infty b_n$converges,and associate$(1)$with M-discriminance we can conclude that $\sum_{n=1}^\infty (a_{n+1}-a_n)$converges.

$2^\circ$For $l-a_n=\sum_{k=n}^\infty (a_{k+1}-a_k)$,associate with$(1)$we get \begin{align*} l-a_n<&\sum_{k=n}^\infty b_k=\sum_{k=n}^\infty \frac{\sqrt{k+1}}{2^k\sqrt{k!}} <\frac{1}{2^n\sqrt{n!}}\sum_{k=n}^\infty\frac{\sqrt{k+1}}{2^{k-n}}\\ =&\frac{1}{2^n\sqrt{n!}}\sum_{k=0}^\infty\frac{\sqrt{n+k+1}}{2^k} =\frac{\sqrt{n}}{2^n\sqrt{n!}}\sum_{k=0}^\infty\frac{1}{2^k}+\frac{1}{2^n\sqrt{n!}}\sum_{k=0}^\infty\frac{\sqrt{n+k+1}-\sqrt{n}}{2^k}\\ =&\frac{2\sqrt{n}}{2^n\sqrt{n!}}+\frac{1}{2^n\sqrt{n!}}\sum_{k=0}^\infty\frac{1}{2^k}\cdot\frac{k+1}{\sqrt{n+k+1}+\sqrt{n}}<\frac{2\sqrt{n}+A}{2^n\sqrt{n!}}~(\text{$A>0$is a constant}) \end{align*} thus \begin{align*} \sqrt{n}\cdot\sqrt[n]{l-a_n}<\sqrt{n}\left(\frac{2\sqrt{n}+A}{2^n\sqrt{n!}}\right)^{\frac1n}\tag{2} \end{align*} By using Stirling's formular,we know that $n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$when $n\to \infty$.Then it's easy to see $\sqrt{n!}\sim \sqrt[4]{2\pi n}\left(\frac{n}{e}\right)^{\frac{n}2}$. So \begin{align*} \lim_{n\to \infty}\sqrt{n}\left(\frac{2\sqrt{n}+A}{2^n\sqrt{n!}}\right)^{\frac1n} =\lim_{n\to \infty}\sqrt{n}\frac{(2\sqrt{n})^\frac1n(1+\frac{A}{2\sqrt{n}})^\frac1n}{2\sqrt[4n]{2\pi n}\left(\frac{n}{e}\right)^{\frac{1}2}}=\frac{\sqrt{e}}{2}\tag{3} \end{align*} Then we let $n\to \infty$and take superior limit in $(2)$,and get \begin{align*} \varlimsup_{n\to \infty}\sqrt{n}\cdot\sqrt[n]{l-a_n}\leqslant \lim_{n\to \infty}\sqrt{n}\left(\frac{2\sqrt{n}+A}{2^n\sqrt{n!}}\right)^{\frac1n}=\frac{\sqrt{e}}{2}\tag{4} \end{align*}

This problem puzzles me is how to prove $\varliminf_{n\to \infty}\sqrt{n}\cdot\sqrt[n]{l-a_n}\geqslant \frac{\sqrt{e}}{2}$. I have tried hard, but without any progress. Can anyone help me?

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