Let $a_i=\sqrt[i]{i+\sqrt[i+1]{i+1+\cdots+\sqrt[n+1]{n+1}}}$, and $b_i=\sqrt[i]{i+\sqrt[i+1]{i+1+\cdots+\sqrt[n]{n}}}$ We will prove by reverse induction that $a_k-b_k<\frac{1}{n\cdot n-1\cdots k}$. First, we will show that: $a_n-b_n<\frac{1}{n}$. $a_n<\frac{1}{n}+\sqrt[n]{n}$ If $n=2$, we have to show that $\sqrt{2+\sqrt[3]{3}}<\frac{1}{2}+\sqrt{2}$, which after squaring becomes $\sqrt[3]{3}<\frac{1}{4}+\sqrt{2}$, and cubing yields $3<\frac{1}{64}+\frac{3}{2}+\sqrt{2}(2+\frac{3}{16})$ so it suffices to show that $\frac{3}{2} < 2\sqrt{2}$, which is true since $\frac{3}{2}<2<2\sqrt{2}$. For $n\ge 3$: Raising both sides to the power of n, and using the binomial theorem, it suffices to show that: $n+\sqrt[n+1]{n+1}<n+n\cdot \frac{1}{n} \cdot \sqrt[n]{n}^{n-1}$ $(n+1)^{\frac{1}{n+1}}<n^{\frac{n-1}{n}}$ $(n+1)^n<n^{(n-1)(n+1)}$ $(n+1)^n<(n^{n-1})^{n+1}$ Which is true since $n^{n-1}>n+1$ and $n+1>n$. Now, assume the result about $a_k-b_k$ holds true for $k=m+1,m+2,...,n$. We will show this implies the result for $k=m$. $a_m-b_m=\frac{a_m^m-b_m^m}{\sum_{i=0}^{m-1}a_m^ib_m^{n-1-i}}=\frac{a_{m-1}-b_{m-1}}{\sum_{i=0}^{m-1}a_m^ib_m^{n-1-i}}$, and since $a_{m-1}-b_{m-1}<\frac{1}{n\cdot n-1\cdots m-1}$, it suffices to prove that $\sum_{i=0}^{m-1}a_m^ib_m^{n-1-i} \ge m$ which is true since it is the sum of $m$ terms, each of which is the product of 2 terms both strictly greater than 1 (and so each term is greater than 1). The original question is this result for $k=2$. Cheers, Rofler

Amir Hossein wrote: Let $x_n = \sqrt[2]{2+\sqrt[3]{3+\cdots+\sqrt[n]{n}}}.$ Prove that \[x_{n+1}-x_n <\frac{1}{n!} \quad n=2,3,\cdots\] Let us note that the estimation is very wide. Though, elegant problem

Attachments:

It seems that \(x_{n+1}-x_{n}<\dfrac{2}{n^n} \) is true as well.

Here a difficult question find an equivalent of x(n)

Rhapsodies_pro wrote: It seems that \(x_{n+1}-x_{n}<\dfrac{2}{n^n}\)（$n\in{\mathbb{N_+}\setminus\{1\}}$） is true as well. However, \[\sqrt{x_{n+1}-x_{n}}<{\prod_{k=2}^{n+1}\frac{\sqrt[2k]{k}}{k-1}}\]seems to be better for sufficiently large \(n\).