# Prove that $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ without calculus?

I'm stuck with this sample RMO question I came across:

Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...

In the solution to this problem, I found the solver making the assumption, $\sqrt[n]{n}>\sqrt[n+1]{n+1}$ for $n \geq 3$ How would you prove this?

Any help would be greatly appreciated.

EDIT: In this competition, you aren't allowed to use calculus. Non-calculus methods would be appreciated.

Assume $n\geq 3.$ Start with this calculation:

$$\frac{(n+1)^n}{n^n} = \left(1+\frac{1}{n}\right)^n = 1+\binom{n}{1}\frac{1}{n} +\binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} +\cdots + \frac{1}{n^n}.$$

In the $k$th term, the numerator of the binomial coefficient is $n(n-1)(n-2)\cdots (n-k+1)$ which is less than $n\cdot n\cdots n=n^k.$ So the binomial expansion above is less than

$$1+1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \frac{1}{n!},$$

which, in turn is less than

$$1+1+ \frac{1}{2}+\frac{1}{2^2} + \cdots +\frac{1}{2^n} < 3\leq n.$$

So we have $$n > \frac{(n+1)^n}{n^n}$$

$$n^{n+1} > (n+1)^n$$

$$n^{(n+1)/n} > n+1$$

$$n^{1/n} > (n+1)^{1/(n+1)}.$$

• Perfect. Thanks a lot! – Abhirup Ghosh Oct 4 '17 at 16:50

consider the function $f(x)=x^{1/x}$ for $x >0$. Now check it's monotonicity. \begin{align*} \ln f(x) & = \frac{\ln x}{x}\\ \frac{f'(x)}{f(x)} & = \frac{1-\ln x}{x^2}\\ f'(x) & = f(x)\left[\frac{1-\ln x}{x^2}\right] \end{align*} Observe that for $x >e$, we get $f'<0$. So for $n \geq 3$, we have $f(n) > f(n+1)$.

• Non-calculus method? – Abhirup Ghosh Oct 4 '17 at 11:42

We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or $$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$ Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.

Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$ for all $x>e$.

Thus, for all $n\geq3$ we have $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ Now, for $n=2$ we get $$\sqrt2<\sqrt[3]3,$$ and for $n=1$ we have $1^1<2^\frac{1}{2}$, which gives that $\sqrt[3]3$ is a largest number in the sequence.

Done!

• How would you prove this? – Abhirup Ghosh Oct 4 '17 at 11:01
• @user2606764 I added something. See now. – Michael Rozenberg Oct 4 '17 at 11:05
• Early comment... Thanks a lot! How would you solve this without using calculus though? – Abhirup Ghosh Oct 4 '17 at 11:07
• @user2606764 You are welcome! – Michael Rozenberg Oct 4 '17 at 11:08
• Just made an edit. – Abhirup Ghosh Oct 4 '17 at 11:09

HINT.-The function $f(x)=x^{\frac 1x}$ has as derivative $f'(x)=-x^{\frac 1x-2}(\ln(x)-1)$ which prove that $f$ is decreasing on $x\gt e$ You can deduce that the maximum value is $\color{red}{\sqrt[3]3\approx1.442249}$ (verifying that $\sqrt2\approx1.414243$).

• Non-calculus method? – Abhirup Ghosh Oct 4 '17 at 11:42
• Prove that $$(3+n)^{\frac{1}{3+n}}\lt3^{\frac13}$$ You have $$3+n\lt3\cdot3^{\frac n3}$$ $$n=1\Rightarrow4\lt4.32....\\n=2\Rightarrow5\lt6.24....\\n=3\Rightarrow6\lt9$$ For $n\gt3$ it is clear. – Piquito Oct 4 '17 at 12:03

We wish to compare $\sqrt[n]n \lessgtr \sqrt[n+1]{n+1}$. Raise each side to the $n(n+1)$th power to get $$n^{n+1} \lessgtr (n+1)^n$$ and use the binomial theorem on the right-hand side: $$n\cdot n^n \lessgtr \underbrace{n^n+\binom n1 n^{n-1} + \binom n2 n^{n-2} + \cdots + \binom n{n-1} n^1}_{n\text{ terms}} + 1$$ Because $\binom{n}{k}\le n^k$, each of the $n$ indicated terms is at most $n^n$. And when $n\ge 3$, the last term $\binom n{n-1}n^1 = n^2$ is so much smaller than $n^n$ that the final $1$ term is insufficient to make the RHS exceed $n\cdot n^n$.