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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 2:56 PM


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.



