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Let $f(t)=\sqrt[t]t=t^\frac 1t=e^{\frac{\ln t}t}$ , $t>0$ , we have

$\displaystyle f'(t)=\frac{e^{\frac{\ln t}t}(1-\ln t)}{t^2}$

Therefore, the largest number of the infinite sequence must be $\sqrt2$ or $\sqrt[3]3\ (e\approx 2.7)$ .

$9>8\implies\sqrt[6]9>\sqrt[6]8$ i.e. $\sqrt[3]3>\sqrt2$

Hence the maximum is $\sqrt[3]3$ , making the answer $2\times3=\boxed 6$ .

Let $f(x) = \sqrt[x] x$ . We note that $\sqrt 2=\sqrt[4] 4$ . From $\dfrac d{dx} f(x) = f'(x) = \left(\dfrac 1{x^2} - \dfrac {\ln x}{x^2}\right) \sqrt [x] x$ , we have $f'(2) > 0$ and $f'(4) < 0$ . This means that $f(x)$ is increasing at $x=2$ and decreasing at $x=4$ . Since $f(x)$ is continuous, its maximum must occur between $2 < x < 4$ . Then for integer $x$ , the maximum value of $\sqrt[x] x$ is at $x=3$ . Therefore, $2x=\boxed{6}$ .

Alternative solution: Let $f(x) = \sqrt[x] x = x^\frac 1x = e^{\frac {\ln x}x}$ . To find the maximum value of $f(x)$ , let us consider $x$ as continuous real instead of discrete integer. Then $\dfrac d{dx} f(x) = f'(x) = \left(\dfrac 1{x^2} - \dfrac {\ln x}{x^2}\right) \sqrt [x] x$ . Putting $f'(x)=0$ , we have $\ln x = 1$ $\implies x = e$ . Note that $f''(x) = \left(- \dfrac 2{x^3}\left(1-\ln x\right) - \dfrac 1{x^3}\right)\sqrt[x] x + \dfrac 1{x^4}\left(1-\ln x\right)^2 \sqrt[x] x$ $\implies f''(e) < 0$ , which means that $f(e) = \sqrt[e] e$ is the maximum value of the continuous function. The nearest integer to $e$ is 3, therefore $x=3$ , when $\sqrt[x] x$ is maximum and $2x = \boxed{6}$ .

Same problem with same solution poster , @Chew-Seong Cheong