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'Discrete Trends' printed from https://nrich.maths.org/
Congratulations to Curt from Reigate College for cracking this
tough nut problem. Here is Curt's solution:
If $n = n^{1\over n}$, I first assert that for an $n> 1$, then
$n^{1\over n}$ is larger than 1
If $n> 1$ then, raising both sides to the power of $1/n$ yields
$n^{1\over n}> 1^{1\over n}$. So $n^{1\over n}$ is larger than 1
no matter what the value of $1/n$. Therefore, one could make
$n^{1\over n}=1+R$. As $n^{1\over n}$ is always larger than 1, $R$
is always a positive real number. Now, raising both sides to the
power $n$ one obtains:
$$n = 1 + nR + \textstyle{1\over 2}n(n-1)R^2 +\cdots.$$
(Incidentally, $n> nR$, therefore $R< 1$ and so $n^{1\over
n}< 2$)
As n is an integer larger than one, at least the third term of
expansion must exist. It is clear that:
$$\eqalign{ n &> \textstyle{1\over 2}n(n-1)R^2 \cr {2n\over
n(n-1)}&> R^2\cr \sqrt{2\over n-1} &> R }.$$
So $R\rightarrow 0$ as $n\rightarrow \infty$ and
$$n^{1\over n} = 1 + R < 1 + \sqrt{2\over n-1}.$$
So $n^{1\over n}\rightarrow 1$ as $n\rightarrow \infty$.
As a corollary, I will show that $\sqrt{2/(n-1)}$ grows smaller for
successive values of $n> 1$. Assume that for some positive $B$,
$\sqrt{2/(n-1)}< \sqrt{2/(n+B-1)}$. Now if this is to be true
$1/(n-1)< 1/(n+b-1)$, therefore $n-1> n+B-1$, therefore
$B< 0$, contrary to our conditions. Therefore the assertion that
for some positive B the value of the function increases is
fallacious. This will prove useful later.
For $n= 1,\ 2,\ $ etc. we have $n= 1,\ 2^{1/2},\ 3^{1/3},\
4^{1/4}=2^{1/2}\ ...$ and we see that the values increase to a
maximum of $3^{1/3}$ and then start to decrease.
Now to prove that 3 is the value of $n$ pertaining to the maximum
of this discrete function, I will find an $n$ such that
$1+\sqrt{2/(n-1)}$ is less that the cube root of 3. From that point
onwards, it is known that for successive values $\sqrt{2/(n-1)}$
decreases, therefore beyond this point the value of $n^{1\over n}$
decreases; more importantly, is less than $3^{1\over 3}$. For
$n=19$ this holds true, as it will for subsequent values. The
"in-between" values have been checked and are less than $3^{1\over
3}$. If $n\geq 19$ then
$$n^{1\over n} < 1 + \sqrt{2\over 18}= {4\over 3}< 3^{1\over
3}$$