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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

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Axel Vogt

Posts: 1,038
Registered: 5/5/07
Re: sum(eulerphi(n)/n!,n=1...infinity) = 5*sqrt(Pi)/3 -1 ????
Posted: Feb 2, 2013 2:56 PM
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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) <= 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.




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