# Prove that $\sqrt[n]n>\sqrt[n+1]{n+1}$ for all $n \geq 3$

Can someone please prove the following?

$$\sqrt[n]n>\sqrt[n+1]{n+1} \quad \text{for all } n\geq 3.$$

I have tried lots of different approaches but none of them has worked. I tried induction and also tried to modify the expression but nothing worked.

• Pretty easy to do by induction, describe what you tried – Alec Teal Nov 1 '14 at 20:18
• I have started with regular induction base case, step case and so on but i just cant find a way to make use of the induction hypothesis – Baquesh Nov 1 '14 at 20:26
• What is your induction hypothesis (you'll do well on this site if you put "what have I tried" (two underscores, title, two underscores, an underscore is a _) (this is a title in your post) then describe your effort, even if it's wrong. – Alec Teal Nov 1 '14 at 20:27
• – Martin Sleziak Oct 4 '17 at 12:06

Hint: Remember that $n \geq 3$. $$\sqrt[n]{n} > \sqrt[n+1]{n+1}\Leftrightarrow n^{n+1} > (n+1)^n$$

One more step,

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

• i tried that way to it didnt work. – Baquesh Nov 1 '14 at 20:24
• @Bakyr Did you get it? – Aaron Maroja Nov 1 '14 at 20:34
• to be honest no. i dont get how you get to this step – Baquesh Nov 1 '14 at 20:35
• As $n\geq 3$ you may divide by $n^{n}$ on both sides of the inequality. Once you have done that, notice that $(1 + \frac{1}{n})^n < 3 ,\forall n$. Remember that $\lim_{n\to\infty} (1 + \frac{1}{n})^n = e < 3$.Then the proof is done. – Aaron Maroja Nov 1 '14 at 20:38
• now i get it. thank you , but i am not only interested for the inequality to hold at infinity, i want it to hold for every n>=3 – Baquesh Nov 1 '14 at 20:42

This can be proved by calculus. Let $y = \sqrt[x]{x} = x^{1/x}$. Then $\displaystyle\ln y = \frac{\ln x}{x}$, so that $$\frac{1}{y} y' = \frac{1-\ln x}{x^2} \implies y' = x^{1/2} \cdot \frac{1-\ln x}{x^2} = \frac{1-\ln x}{x^{3/2}}.$$ Since $1-\ln x < 0$ for all $x \geq 3$, we have $y'<0$ on $[3,\infty)$, that is, $y$ is a decreasing function on $[3,\infty)$.

• what does that y' mean – Baquesh Nov 1 '14 at 20:33
• $y'$ is the derivative of $y$ with respect to $x$, that is, $\displaystyle\frac{dy}{dx}$. – E W H Lee Nov 1 '14 at 20:35