Greatest value of $n^{1/n}$ without calculus where n is a natural number.

Question

I want to find the greatest value of $n^{1/n}$ where n is a natural number without using calculus

As n tends to infinity $n^{1/n}$ will get very small.
$3^{1/3}$ looks like the largest because $1^1<2^{1/2}<3^{1/3}>4^{1/4}$
The only thing I could think of was using binomial theorem to prove that
${(n)}^{1/n}>(n+1)^{1/{(n+1)}}$ for $n \ge 3$
However, after applying the binomial theorem, I am not able to prove the inequality.
Is using binomial theorem, not the right approach? What could be the other approach?
Can anybody help?

Answer 1

You can use the binomial theorem but in a slightly adopted manner.

Consider

$$\left(\frac{(n+1)^{\frac 1{n+1}}}{n^{\frac 1n}}\right)^{n(n+1)}=\frac 1n\left(1+\frac 1n\right)^n$$

Now use the binomial theorem on $\left(1+\frac 1n\right)^n$ and note that for $n\geq 3$ and $2\leq k\leq n$ you have

$$\binom{n}{k}\frac 1{n^k}<\frac 1{k!}\leq\frac 1{2^{k-1}}$$ Summing up the binomial expansion and using these estimates, you get for $n\geq 3$

$$\left(1+\frac 1n\right)^n<3 \stackrel{n\geq 3}{\Rightarrow} \frac 1n \left(1+\frac 1n\right)^n<1$$

Checking the cases $n=1$ and $n=2$ you get $\boxed{n=3}$ as maximum.

Answer 2

Notice that $n^{1/n} > 1$ for all $n \in \mathbb N$.

Raise both sides of $n^{1/n} \stackrel?> (n+1)^{1/(n+1)}$ to the $n(n+1)$-th power, now we just need to show that

$$n^{n+1} > (n+1)^n \text{ for } n \ge 3$$

which can be shown by induction quite easily.

See Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$.