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need help about 2 constants

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Pierre CAMI     Message 1 of 3  Mar 10, 2006
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I try to calculate the constant sum from 1 to infinity
of n/(p(n)^2) with p(n) = n-th prime as i d'nt found this value 
searching on the net.
It seems to me that this need to converge as for high n it is
like n/(p(n)^2) ~ 1/n*(log(n)^2) and n*(log(n)^2) > n^1
I found that it is > 1.1 with a very low convergence , so is it a way 
to found this value with let say 5 or 10 digits after the 1 ?
Same question about sum of 1/n*(log(n)^2)
Thanks for any help 
Pierre
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David Broadhurst     Message 2 of 3  Mar 10, 2006
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--- In primeform@yahoogroups.com, "Pierre CAMI" <pierrecami@...> 
wrote:

> Same question about sum of 1/n*(log(n)^2)

I assume that you mean the sum of 1/(n*log(n)^2) 
over all integers n>1.

Here you can use an Euler-MacLaurin
method since you are have a smooth (infinitely differentiable)
function summed over integers and the integral of 1/(x*log(x)^2) 
is -1/log(x), which vanishes at infinity. 

Using only a single derivative I got

2.10974280123689221
2.10974280123689138
2.10974280123688166

by truncating at n = 10^3, 10^4, 10^5.

This can easily be improved by taking the third derivative:

2.1097428012368919744790
2.1097428012368919744792
2.1097428012368919744792

BUT, for the sum of n/prime(n)^2 
you have no hope of using the Euler-MacLaurin method, 
since the fluctutations of primality destroy the
assumption of smoothness

So you asked one very hard question 
and one very easy question.

David
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David Broadhurst     Message 3 of 3  Mar 10, 2006
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--- In primeform@yahoogroups.com, "David Broadhurst"
<d.broadhurst@...> wrote:

> > Same question about sum of 1/n*(log(n)^2)
.. 
> 2.1097428012368919744790
> 2.1097428012368919744792
> 2.1097428012368919744792

PS: If you Google

2.10974280123689

you will see that this constant appears in a college
mathematics journal to which I do not have access.

>>
MR1478271 
Kreminski, Rick 
Using Simpson's rule to approximate sums of infinite series. 
College Math. J. 28 (1997), no. 5, 368--376.
65B10
[There will be no review of this item.]
<<

Perhaps college students know about 
Bart Simpson, but not about Leonhard Euler :-?

Here is a more accurate value, courtesy of Leonhard:

2.1097428012368919744792571976165513263855319843947420226499156

As you can see analysis is quite easy; 
it is /arithmetic/ that is truly hard.

David
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