Evaluate $$\sqrt{e\sqrt{e\sqrt{e\sqrt{e\cdots}}}}$$
My attempt:
Let $$x=\sqrt{e\sqrt{e\sqrt{e\sqrt{e\cdots}}}}$$
$$x^2=e\sqrt{e\sqrt{e\sqrt{e\cdots}}}$$
But I've no idea, how to proceed. Hope someone can point it out. Thanks.
$\endgroup$ 44 Answers
$\begingroup$Consider $\{x_n\}$: $x_{n+1}^2=ex_n$, $x_1=\sqrt{e}$.
Since $x_1<e$ and $$x_{n+1}-e=\frac{\sqrt{e}(x_n-e)}{\sqrt{x_n}+e}$$ by induction we get $x_n<e$ and $$x_{n+1}-x_n=\sqrt{x_n}(\sqrt{e}-\sqrt{x_n})>0.$$
Thus, there is $\lim\limits_{n\rightarrow+\infty}x_n$.
Let $\lim\limits_{n\rightarrow+\infty}x_n=x$.
Thus, $x^2=ex$, which gives $x=e$.
$\endgroup$ 1 $\begingroup$Let $$x=\sqrt{e\sqrt{e\sqrt{e\sqrt{e\cdots}}}}$$ Then $$x=\sqrt{ex}$$ $$\Downarrow$$ $$x^2=ex$$ $$\Downarrow$$ $$x=e$$
$\endgroup$ $\begingroup$$\frac{x^2}{e} = x$, $ x > 0 $ so $ x=e$
$\endgroup$ $\begingroup$In this case you have $$ x^2 = ex $$ and you can solve this for $x$.
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