Inegalitate

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alex2008
Leibniz
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Joined: Sun Oct 19, 2008 3:23 pm
Location: Tulcea

Inegalitate

Post by alex2008 »

Sa se arate ca \( (1+\frac{1}{n})^n\le (1+\frac{1}{n+1})^{n+1} ,\ (\forall)n\ge 1 \) .
. A snake that slithers on the ground can only dream of flying through the air.
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Marius Mainea
Gauss
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Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Flosim Inegalitatea lui Bernoulli:

\( (1+x)^n\ge 1+nx \) pentru orice \( x\ge-1 \)

Astfel dupa putine calcule inegalitatea noastra se reduce la

\( \frac{n+1}{n+2}\le [1-\frac{1}{(n+1)^2}]^n \)

care este adevarata luind in inegalitatea lui Bernoulli \( x=-\frac{1}{(n+1)^2} \)
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