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Found this inside an example in my analysis book. Can't figure out why the two are equivalent and how to work out the algebra.

My question is:

  • What is the algebra behind $n^{1/n}>(n+1)^{1/(n+1)}$ is equivalent to $n^{n+1}>(n+1)^n$ (and how to prove it)?

Sorry if this is something basic, I'm just not being able to work out the algebra behind it.

EDIT: I think I need to take a break...

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$$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ it's $$\left(n^{\frac{1}{n}}\right)^{n(n+1)}>\left((n+1)^{\frac{1}{n+1}}\right)^{n(n+1)}$$ or $$n^{n+1}>(n+1)^{n}$$

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$f(x)=x^{n(n+1)}$ is strictly increasing for $x\geq 0$.

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  • $\begingroup$ $f(x)=x^6$ is strictly increasing ? $\endgroup$ – Aqua Jan 7 '18 at 17:48
  • $\begingroup$ It means $x\geq0$. $\endgroup$ – Michael Rozenberg Jan 7 '18 at 17:50
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powering by $n(n+1)$ we get $$n^{n+1}>(n+1)^n$$ and it follows that $$\left(1+\frac{1}{n}\right)^n<n$$ for $n\geq 2$

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