# Continuous of square root of e [duplicate]

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.

• It looks like it's the same as $x^2 = ex$ by definition of $x$ – Brenton Jul 22 '17 at 1:45
• @Mathxx $x^2=ex$. – Michael Rozenberg Jul 22 '17 at 1:45
• You should really first define what an infinite nested square root means - probably a limit of $n$ nested square roots as $n$ goes to infinity. Then, you should prove that this limit exists and has a finite value. At this point, since you know that the infinite nested square root converges, you can let its value be $x$, and manipulate it to get $x^2 = ex$ as others have mentioned. Technically, you cannot let the value be $x$ until you are sure it converges. – Zubin Mukerjee Jul 22 '17 at 2:10
• The expression is equal to $$\exp\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots\right)$$ which is same as $\exp(1)=e$ because the infinite series is a GP which adds up to $1$. – Paramanand Singh Jul 22 '17 at 6:50

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 $x=e$.

• This is more complete than the other answers, since it defines a sequence whose limit is the infinite nested square root, and shows that the limit exists. Very nice, +1! – Zubin Mukerjee Jul 22 '17 at 2:12

Let $$x=\sqrt{e\sqrt{e\sqrt{e\sqrt{e\cdots}}}}$$ Then $$x=\sqrt{ex}$$ $$\Downarrow$$ $$x^2=ex$$ $$\Downarrow$$ $$x=e$$

$\frac{x^2}{e} = x$, $x > 0$ so $x=e$

In this case you have $$x^2 = ex$$ and you can solve this for $x$.