# Proving $n^{1/n}$ is Monotonically Decreasing

1. Jan 4, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
I am trying to show that $n^{1/n}$ is monotonically decreasing for $n \ge 3$.

2. Relevant equations

3. The attempt at a solution
I am trying to prove the claim using induction. The base case is involves a trivial calculation. What I am having trouble is the induction step; i.e., assuming that $n^{1/n} > (n+1)^{\frac{1}{n+1}}$ is true, I want to show $(n+1)^{\frac{1}{n+1}} > (n+2)^{\frac{1}{n+2}}$.

I have written the original inequality as $n^{n+1} > (n+1)^n$ and $n > \left(\frac{n+1}{n} \right)^n$ and working with these, but i have had no success. I could use a hint.

2. Jan 4, 2017

### mjc123

Would it be cheating to differentiate the continuous function x1/x? If (over a certain value of x) the differential is always negative, the function is monotonically decreasing, therefore so are its values for integral values of x.

3. Jan 4, 2017

### Logical Dog

the induction step starts with a(3) > a(4)

does n mean that the fucntions domain is natural numbers/

4. Jan 4, 2017

### Bashyboy

I would like to avoid using derivatives and other calculus techniques

Yes, $n$ denotes a natural number. I am trying to show that the sequence $n^{1/n}$ converges to $1$.

5. Jan 4, 2017

### Logical Dog

1/n converges to 0 as n tends to infinty and any natural number to the power 0 is 1 (this not a proof but intuition)
you want to prove monotonicity though.

Its enough to show that the function is bounded and convergent and hence has a monotonic subsequence, but I am sure this statement is not true

i think induction is the only way, i tried it and have got nowhere :P g

Last edited: Jan 4, 2017
6. Jan 4, 2017

### PeroK

First, if you are trying to show that the limit is 1, then it will not be enough to show that the sequence is monotonic.

Second, induction looks out of place here, as each inequality does not appear to depend on the previous one.

A hint for the limit is to consider the sequence $a_n = n^{1/n} - 1$ and try to show that this converges to 0.

7. Jan 4, 2017

### Bashyboy

It can be done. See Eric Naslund's answer and Martin Sleziak's comment on Eric's comment: http://math.stackexchange.com/questions/76330/prove-sequence-a-n-n1-n-is-convergent

I tried to understand Sleziak's approach to induction, but I couldn't quite follow it. I understand he is proving it by contradiction; but if it really were a proof by contradiction, why wouldn't he end up with the inequalities like $n^{1/n} > (n+1)^{\frac{1}{n+1}}$ and $(n+1)^{\frac{1}{n+1}} \le (n+2)^{\frac{1}{n+2}}$ rather than $(n-1)^n\ge n^{n-1}$ and $(n+1)^n> n^{n+1}$?

8. Jan 4, 2017

### Bashyboy

Bipolar Demon: There may be a way to boost this intuition to a proof. We can write

$$\lim_{n \rightarrow \infty} = n^{1/n} = \lim_{n \rightarrow \infty} e^{\ln n^{1/n}} = e^{\lim_{n \rightarrow \infty} \frac{\ln n}{n}}$$

where the last step invoked the composition limit law. However, in order to justly use this law, we need to show that $\lim_{n \rightarrow \infty} \frac{\ln n}{n}$ exists and, in particular, is zero.

I would like to see both of these proofs work.

9. Jan 4, 2017

### PeroK

Looks like a mess to me! Instead try the method I suggested and the next step/hint is:

$(1 + a_n )^n = n$

10. Jan 4, 2017

### PeroK

If you assume that $\lim_{n \rightarrow \infty} \frac{\log n}{n} = 0$, then, with $b_n = n^{1/n}$ we have:

$\lim_{n \rightarrow \infty} \log b_n = \lim_{n \rightarrow \infty} \frac{\log n}{n} = 0$

And, as $\log$ is a 1-1 function, this can only happen if $\lim_{n \rightarrow \infty} b_n = 1$

11. Jan 4, 2017

### Bashyboy

Hmmm...Should I be doing an $\epsilon$-proof with this hint?

12. Jan 4, 2017

### PeroK

There's no need to get into $\epsilon-\delta$. Instead, expand the power using the Binomial theorem and see if anything catches your eye.

That's three hints now!

13. Jan 4, 2017

### Ray Vickson

You have shown that $n^{1/n}$ is decreasing if and only if
$$\left( 1 + \frac{1}{n} \right)^n < n$$.
The binomial expansion of $(1 + 1/n)^n$ gives
$$\left( 1 + \frac{1}{n} \right)^n \leq 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}.$$
Go on from there.