Discrete Trends
Find the maximum value of n to the power 1/n and prove that it is a
maximum.
Age
16 to 18
Challenge level
Show that if $n$ is a positive integer then
$$n^{1/n} < 1 + \sqrt {{2\over {n-1}}}.$$
Show that $n^{1/n}\rightarrow 1$ as $n\rightarrow \infty$.
Find the maximum value of $n^{1/n}$ and prove that it is indeed the maximum.
As $n$ is an integer try finding $n^{1\over n}$ for some small values of $n$. What do you find? If you think you might have found the maximum value then you'll need to use the first part of the question to prove it really is the maximum. As the problem is about discrete (whole number) values you can find a solution without calculus. To show that
$$n^{1/n} < 1 + \sqrt {{2\over {n-1}}}$$
write $n^{1/n} = 1 + \delta$ and use the Binomial Theorem. If $n> 1$ then $\delta> 0$. Throw away all but one term of the Binomial expansion to get the inequality.
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
$$n^{1\over n} < 1 + \sqrt{2\over 18}= {4\over 3}< 3^{1\over 3}$$
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
$$n^{1\over n} < 1 + \sqrt{2\over 18}= {4\over 3}< 3^{1\over 3}$$
Why do this problem?
It gives practice in working with inequalities.
As we know $n$ is a positive integer learners can investigate $n^{{1\over n}}$ for for different values of $n$ and make conjectures about where the maximum value occurs.
Possible approach
You need to find a local maximum for a small value of $n$ and then prove that this is the only maximum value. Clearly it is impossible to check all values of $n$. One method of proving the result uses the Binomial theorem.