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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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jacob navia

Posts: 23
Registered: 4/13/10
Re: sum(eulerphi(n)/n!,n=1...infinity) = 5*sqrt(Pi)/3 -1 ????
Posted: Feb 2, 2013 5:03 PM
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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) <= 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.

(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
---------------------------------------------
-----
\
\
/ (n-1)!
/
-----
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.




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