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I believe that Ln(n) is <= 2^(sqrt(2*lg(n))), so I am trying to prove it

I know that you can show Big O by showing Little O, so I am trying to take the limit as n approaches infinity of Ln(n)/2^(sqrt(2*lg(n)))

However I am finding it hard to prove this using the limit definition

I would appreciate any help to see how to solve this! Thanks

Kneelac
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1 Answers1

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Interpreting lg(n) as log2(n), set n=2^x so that the inequality transforms to

ln(2)*x <= 2^sqrt(2*x)

Now apply the square root,

sqrt(2*ln(2)) * sqrt(x/2) <= 2^sqrt(x/2)

and we are left to prove that for y>0

2^y/y > sqrt(2*ln(2))

Indeed, the minimum of the left side can be found for 0=ln(2)*y-1 or y=1/ln(2) with minimum value 2^(1/ln(2))*ln(2). Numerically, the gap between 1.88416938536372 and 1.1774100225154747 is large enough to exclude any floating point errors. Thus all the equivalent inequalities including the original one are true.

Lutz Lehmann
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