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user21820
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Prove that the nth root of n is greater than (n+1)th root of ($\sqrt[n]{n} > \sqrt[n+1]{n+1)}$ without calculus?

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Martin Sleziak
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I'm stuck with this sample RMO question I came across:

Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...

In the solution to this problem, I found the solver making the assumption, $\sqrt[n]{n}>\sqrt[n+1]{n+1}$. for $n \geq 3$ How would you prove this?

Any help would be greatly appreciated.

EDIT: In this competition, you aren't allowed to use calculus. Non-calculus methods would be appreciated.

I'm stuck with this sample RMO question I came across:

Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...

In the solution to this problem, I found the solver making the assumption, $\sqrt[n]{n}>\sqrt[n+1]{n+1}$. How would you prove this?

Any help would be greatly appreciated.

EDIT: In this competition, you aren't allowed to use calculus. Non-calculus methods would be appreciated.

I'm stuck with this sample RMO question I came across:

Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...

In the solution to this problem, I found the solver making the assumption, $\sqrt[n]{n}>\sqrt[n+1]{n+1}$ for $n \geq 3$ How would you prove this?

Any help would be greatly appreciated.

EDIT: In this competition, you aren't allowed to use calculus. Non-calculus methods would be appreciated.

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