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.