TITLE: Which is greatest in the sequence $1, \sqrt 2 , \sqrt[3] 3 , \sqrt[4] 4 \cdot \cdot \cdot?$
QUESTION [7 upvotes]: 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 $n.$

REPLY [2 votes]: 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 $n \ge 3$.