Which Is The Biggest?

$\sqrt{1} = 1$ . But what can we say about $\sqrt[3]{2}$ versus $\sqrt[4]{3}$ ? They are certainly both larger than 1, but we can use the order-preserving operation $x^{12}$ on each to see which is larger. $(\sqrt[3]{2})^{12} = 16 \text{ and } (\sqrt[4]{3})^{12} = 27.$

Therefore, $\sqrt[4]{3}$ is larger than $\sqrt[3]{2}$ . Now we compare $\sqrt[4]{3}$ with $\sqrt[5]{4}$ . Raising both of these to the 20th power, we get $(\sqrt[4]{3})^{20} = 243 \text{ and } (\sqrt[5]{4})^{20} = 256.$

Therefore, $\sqrt[5]{4}$ is larger than $\sqrt[4]{3}$ . Lastly, we'll compare $\sqrt[5]{4}$ with $\sqrt[6]{5}$ . Raising both of these to the 30th power, we obtain $(\sqrt[5]{4})^{30} = 4096 \text{ and } (\sqrt[6]{5})^{20} = 3125.$

We can conclude that $\sqrt[5]{4}$ is the larger than $\sqrt[6]{5}$ and all numbers after it, since the sequence $\sqrt[n+1]{n}$ approaches 1 as $n \rightarrow \infty$ , meaning the sequence peaks at $n = 4$ and then approaches 1 asymptotically.

To be sure that the sequence doesn't peak again, we have to show the derivative of $f(x)= x^{\frac{1}{x+1}}$ only has one solution. Taking the logarithmic differentiation, we have $\dfrac{f'(x)}{f(x)} = \dfrac{x+ 1-x\ln(x)}{x(x+1)^2}$ . At its extremal point, $f'(x) = 0$ , which means that $x+1 - x\ln(x) = 0$ or $x (\ln(x) - 1) = 1$ . Sketching out the two graphs for $g(x) = \frac1x$ and $h(x) = \ln(x) - 1$ shows that there's only one intersection point. Hence, there's only one solution for $f'(x) = 0$ , equivalently the function $f(x)$ only peak once. In other words, the sequence is decreasing from here on out.