I didn't believe the original posting and I couldn't believe this reply! Have you ever heard about the asymptotic pi(n)~n/ln(n)? How it can be smaller than n^e for e>0? It is greater than n^e for any e<1! Alec Mihailovs http://webpages.shepherd.edu/amihailo/ PS That may be the reason that Edwin Clark and some other people just unsubscribed from this list, because it becomes less and less math and it doesn't seem fun at all? - Alec

-----Original Message----- From: math-fun-bounces+alec=mihailovs.com@mailman.xmission.com [mailto:math-fun-bounces+alec=mihailovs.com@mailman.xmission.com] On Behalf Of Jud McCranie Sent: Saturday, January 24, 2004 7:35 PM To: math-fun Subject: Re: [math-fun] n^e and pi(n)

At 07:05 AM 1/21/2004, you wrote:

...

Is this true?

for any e>0, n^e>pi(n) for all n>N_e

where pi(n) is the number of primes <=n

Yes, n^e grows much faster than pi(n).

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