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Possible Duplicate:
Proving $(1 + 1/n)^{n+1} \gt e$

How to prove this:

$$ \left(\frac{x}{x-1}\right)^x \geq e \qquad\text{for}\qquad x \in \mathbb{N}^* $$

$e$ is the base of the natural logarithm.

and I think the equal satisfies if $x$ converges to infinity.

Thank you!

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    [Related](http://math.stackexchange.com/q/121076/7850) (or duplicate??)2012-05-07
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    Possibly helpful: $$\left(\frac{x}{x-1}\right)^x=\left(\frac{x^2+x}{x^2-1}\right)^x=\left(\frac{x^2}{x^2-1}\right)^x\left(\frac{x^2+x}{x^2}\right)^x=\left(\frac{x^2}{x^2-1}\right)^x\left(1+\frac{1}{x}\right)^x$$2012-05-07
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    Thanks. I should search more thoroughly before posting.2012-05-07
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    Yes, but only to a point. If the linked question were not in the list of my "answers", I might have never found it!2012-05-07
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    Don't the other question deal with $n$ natural, while this deals with $x$ real? Of course, the answers there can probably be modified...2012-05-07

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