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Topic:
sum(eulerphi(n)/n!,n=1...infinity) = 5*sqrt(Pi)/3 1 ????
Replies:
6
Last Post:
Feb 2, 2013 5:14 PM




Re: sum(eulerphi(n)/n!,n=1...infinity) = 5*sqrt(Pi)/3 1 ????
Posted:
Feb 2, 2013 5:03 PM


Le 02/02/13 20:56, Axel Vogt a écrit : > On 02.02.2013 16:34, apovolot@gmail.com wrote: >> Hi, >> >> How close gets >> sum(eulerphi(n)/n!,n=1...infinity) >> to >> 5*sqrt(Pi)/3 1 >> ???? >> >> Regular (free that is) WolframAlpha gives only few decimal digits via >> partial sums ("show points" option) and times out without direct >> evaluation of the sum ... > > > Note that phi(k) <= k1 (and equality for prime numers), almost by > definition. > > For any positive integer we have Sum(phi(k)/k!, k=1 .. infinity) > <= Sum(phi(k)/k!, k=1 .. k0) + Sum(phi(k)/k!, k=k0+1 .. infinity) = > = Sum(phi(k)/k!, k=1 .. k0) + 1/k0! > > For the last evaluation one can use Maple or any other system to see that. > > Now for k0 = 9 that can be computed quickly as 709099/362880 and that is > strictly smaller than your other term by ~ 1e5 > > And for k0 = 20 that error is very small, since 1/k0! > 0 quite fast. >
The Error Term of the Summatory Euler Phi Function N. A. Carella, August 2012.
(http://arxiv.org/pdf/1206.2792.pdf)
There, it is demonstrated that
 \ \ eulerphi(n) 6 /   x + O(1) / n pi**2  n <= x
So we have a series division here since the series proposed by apovolot@gmail.com is
 \ \ eulerphi(n) /  / n  n <= x   \ \ / (n1)! /  n <= x
My math doesn't follow here, but I think it would be possible to do that division isn't it?
In other words we know that even if eulerphi(x)/x doesn't converge eulerphi(x)/n! does.



