Determine the largest number in the infinite sequence
$1,\sqrt{2},\sqrt[3]{3},\sqrt[4]{4},....\sqrt[n]{n}$.
I found $3^{\frac{1}{3}}$ is largest.
I am confused how to start the question. Please help me
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4http://math.stackexchange.com/questions/116112/find-the-maximum-of-fx-x1-x – lab bhattacharjee Dec 20 '16 at 09:14
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3consider the function $$f(x)=x^{1/x}$$ – Dr. Sonnhard Graubner Dec 20 '16 at 09:18
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2Hint: $\sqrt[e]{e}$ is the largest. BTW, this question has been asked here approximately $200$ times. – barak manos Dec 20 '16 at 09:21
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Are you considering $n\in \mathbb N$? – Dec 20 '16 at 09:22
2 Answers
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Let $f(x)=x^{1/x}$ for $x \ge 1$. Show, by using the derivative, that $f$ attaines a global maximum at $x=e$. Hence $f$ is increasing on $[1,e]$ and decreasing on $[e, \infty)$. Since $2<e<3$ amd $1 <\sqrt{2}<\sqrt[3]{3}$ we have $\sqrt[3]{3}>\sqrt[n]{n}$ for $n \ge 4$
Fred
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If we find the derivative of the continuous function $y=x^\frac 1 x$ we get:
\begin{align} \ln y&=\frac 1 x \ln x\\ \frac 1 y \frac{\text dy}{\text dx}&=\frac{1-\ln x}{x^2}\\ \frac{\text dy}{\text dx}&=\frac{x^\frac 1 x(1 - \ln x)}{x^2} \end{align}
Setting this to zero, we get:
$$1-\ln x = 0\\x=e$$
Since you have presented a discrete function, $e$ is not in the domain. However, $e$ is closest to 3.
Richard Ambler
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