Which is greatest in the sequence $1, \sqrt 2 , \sqrt[3] 3 , \sqrt[4] 4 \cdot \cdot \cdot?$ This problem is from Ivan Niven's "Maxima and Minima Without Calculus". What is another way to find this? The solution from the book was:
Note that $\large \sqrt[4] 4 =\sqrt 2$ so this hints that $\sqrt[3] 3$ is the largest. Next the book proved that $\sqrt[3] 3>n^{1/n}$ or $3^n>n^3$ for large enough $n.$
 A: Hint 
The function $x\mapsto x^n$ is monotonically increasing on $[0,+\infty)$.
For example to compare $\sqrt[3] 3 $ and $\sqrt[4] 4 $ we compare $(\sqrt[3] 3)^{12}$ and $(\sqrt[4] 4 )^{12}$.
A: Hmmm, not sure how to do with pre-calculus knowledge, but with calculus, I'd show:

$f(x)=x^{1/x}$ has a maximum when $x=e$ and is increasing when $x<e$ and decreasing when $x>e$.

So it has to be either $f(2)$ or $f(3)$. 
Then check those two yourself - it's pretty easy.
A: Try taking $\sqrt 2$ and $\sqrt[3]3$ to the sixth power. Which is larger? Since $x\mapsto x^6$ is an increasing function on the non-negative reals, what can you conclude?
A: Comparing
$n^{1/n} ? (n+1)^{1/(n+1)}$
(i.e., we want to find out
if "?" is ">" or "<").
Raise to the
$n(n+1)$ power:
$n^{n+1} ? (n+1)^n$.
Divide by $n^n$:
$n ? (1+1/n)^n$.
Apply one of the many elementary ways
to show that
$(1+1/n)^n <e < 3$
such as this:
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?.
Therefore,
"?" is ">" for
$n \ge 3$,
so
$n^{1/n} > (n+1)^{1/(n+1)}$
for $n \ge 3$.
A: If $f(x)=x^{1/x}$ then $f'(x)=0$ only at $x=e$. So you have to check only $2$ and $3$.
