Le 02/02/13 20:56, Axel Vogt a écrit : > On 02.02.2013 16:34, firstname.lastname@example.org 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) <= k-1 (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 ~ 1e-5 > > 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.