TITLE: if $x_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}$ prove
QUESTION [4 upvotes]: if $x_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}$ prove that for $n\ge 2$ $$x_{n+1}-x_n<\frac{1}{n!}$$.

I think induction works best here. The case when $n=2$ is easy as $x_3-x_2=\sqrt{2+\sqrt[3]{3}}-\sqrt{2}=0.44<\frac{1}{2!}$.
However i am not able to proceed further ,the complex radicals are creating a lot of trouble

REPLY [5 votes]: Hint: use the fact that $$\sqrt{y}-\sqrt{x}=\frac{y-x}{\sqrt{y}+\sqrt{x}}$$
For example
$$x_{n+1}-x_n=\frac{\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}-\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}{\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}}+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}}<\\
\frac{\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}-\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}{2}<...$$
Next $$\sqrt[3]{y}-\sqrt[3]{x}=\frac{y-x}{\sqrt[3]{y^2}+\sqrt[3]{yx}+\sqrt[3]{x}}$$
leading to
$$...<\frac{1}{2}\frac{\sqrt[4]{4+...\sqrt[n+1]{n+1}}-\sqrt[4]{4+...\sqrt[n]{n}}}{3}$$
And so on ...

General idea comes from the following identity $$y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+y^{n-3}x^2+...+yx^{n-2}+x^{n-1})$$
Replace $y\to \sqrt[n]{y}$, $x\to \sqrt[n]{x}$ and consider that both $y>1$ and $x>1$. As a result
$$\sqrt[n]{y}-\sqrt[n]{x}<\frac{y-x}{n}$$
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