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October 19th, 2010, 06:59 AM   #1
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Asymptotics

Hello. Here is the problem:

. How to find where .

I need your ideas Thanks
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October 19th, 2010, 08:38 AM   #2
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Re: Asymptotics

Well, k! is about (k/e)^k, so 1/log(k) is about 1/(k log k - k). integral 1/k log k up to m is log log m, so that's a reasonable guess. Actually integral 1/(k log k - k) is log(log m - 1) so that 'should be' even more accurate.
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