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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 11:33 AM
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Hi

Here is a C program for you
----------------------------------------------------------cut here
/*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 ...

Regards,
ARP
*/

/* Compiled with the free compiler lcc-win32. Download that compiler from
http://www.cs.virginia.edu/~lcc-win32.

It is a free C compiler with extended floats of 100 digits precision in the
32 bit version, 130 in the 64 bit version
*/

#include <qfloat.h>
/* returns Euler's totient phi function */
qfloat eulerphi (int n)
{
qfloat phi = 1;
int p;

for (p = 2; p * p <= n; p += 2)
{
if (n % p == 0)
{
phi *= p - 1;
n /= p;
while (n % p == 0)
{
phi *= p;
n /= p;
}
}

if (p == 2)
p--;
}

/* now n is prime or 1 */

return (n == 1) ? phi : phi * (n - 1);
}

int main(void)
{
qfloat sum=0;
qfloat factorial=1;
qfloat pi = 4.0Q * atan(1.0Q);
qfloat delta;

for (int i=1; i<= 10000; i++) {
factorial = factorial * i;
sum += eulerphi(i)/factorial;
}
printf("sum:\n%80.75qg\n",sum);
printf("formula:\n%80.75qg\n",5.0Q * sqrt(pi)/3.0Q - 1.0Q);
delta = sum - (5.0Q * sqrt(pi)/3.0Q - 1.0Q);
printf("delta is:\n%80.75qe\n",delta);
}

The output for a run with 10000 points is:

sum:
1.95408535787600621314459486114714954037067277018696106082987772078870879855276092
formula:
1.95408975150919337883027913890190863799591576020397854702301298308818547431838697
delta is:
-0.00000439363318716568568427775475909762524299001701748619313526229947667576562604516778

BEWARE:

I may have a bug in my eulerphi function that I adapted from an integer
version.I verified the first 100 only.

jacob



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