Efficient way to compare roots (calculator doesn't count)

Question

How can I know which one of the following numbers is the greatest: $$2^{1/2},3^{1/3},4^{1/4},5^{1/5},{6^{1/6}}$$ That can be also written as: $$\sqrt[2]{2},\sqrt[3]{3},\sqrt[4]{4},\sqrt[5]{5},\sqrt[6]{6}$$ Is there a fast way to determine what the greatest number is without a calculator? $n$th root algorithm is not fast so I can't use it. How can we solve this and similiar problems only by hand?

Answer 1

$$\begin{array}{rrcl} & n^{1/n} &>& (n+1)^{1/(n+1)} \\ \iff& n^{n+1} &>& (n+1)^n \\ \iff& n &>& \displaystyle \left(1+\frac1n\right)^n \\ \impliedby& n &>& e \end{array}$$

Since $3 > e$, we can be sure that $3^{1/3} > 4^{1/4} > 5^{1/5} > 6^{1/6}$.

As for $2$, notice that $\left(1+\dfrac12\right)^2 = 2.25 > 2$, so $2^{1/2} < 3^{1/3}$.

So the greatest number is $3^{1/3}$.

Answer 2

Their logarithms are $$\frac{\ln n}{n}$$ for $n=2,\ldots,6$. The function $f(x)=(\ln x)/x$ is increasing on $(1,e)$, has a maximum at $e$, and is decreasing on $(e,\infty)$. To find the largest in your list, just compare $f(2)$ and $f(3)$. To find the smallest in your list, just compare $f(2)$ and $f(6)$.

Answer 3

Hint: The function $f(x)=x^{1/x}$ is decreasing for $x \ge e$. It only remains to compare $2^{1/2}$ and $3^{1/3}$. For that, raise them to the $6$-th power.

Answer 4

To compare fractions, you would put them over the same denominator, and you could apply a similar method in this case, using $(x^a)^b=x^{a b}$.

For example, you can write $2^{1/2} = (2^3)^{1/6} = 8^{1/6}$ and $3^{1/3} = (3^2)^{1/6} = 9^{1/6}$, which makes it clear that $3^{1/3} > 2^{1/2}$.

You can compare all five in one go by writing them all in the form $n^{1/60}$, but whether you can calculate, e.g., $5^{12}$ without a calculator is another matter.