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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.

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

One more step,

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

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  • $\begingroup$ i tried that way to it didnt work. $\endgroup$ – Baquesh Nov 1 '14 at 20:24
  • $\begingroup$ @Bakyr Did you get it? $\endgroup$ – Aaron Maroja Nov 1 '14 at 20:34
  • $\begingroup$ to be honest no. i dont get how you get to this step $\endgroup$ – Baquesh Nov 1 '14 at 20:35
  • $\begingroup$ 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. $\endgroup$ – Aaron Maroja Nov 1 '14 at 20:38
  • $\begingroup$ 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 $\endgroup$ – Baquesh Nov 1 '14 at 20:42
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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 $[3,\infty)$.

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

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