Any idea how to find at least one "n" such that:

[ ( x^{1/n}+ y^{1/n}) / 2 ]^{2n }- xy < epsilon

2 <= x <= a, 2 <= y <= a

for any epsilon?

Actually, its well-known that:

lim [ ( x^{1/n}+ y^{1/n}) / 2 ]^{2n = xy}

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- November 14th 2012, 09:09 AMymsn-epsilon relationship for geometric mean limit
Any idea how to find at least one "n" such that:

[ ( x^{1/n}+ y^{1/n}) / 2 ]^{2n }- xy < epsilon

2 <= x <= a, 2 <= y <= a

for any epsilon?

Actually, its well-known that:

lim [ ( x^{1/n}+ y^{1/n}) / 2 ]^{2n = xy} - November 14th 2012, 05:28 PMymsRe: n-epsilon relationship for geometric mean limit
Even for lim n^(1/n) = 1 it is a problem ...

(1/n) (log n) < log(1 + eps) , n > ?? ...