Date: Wed, 27 Mar 96 13:26:10 CST
From: rusin (Dave Rusin)
To: rdsilverman@qed.com
Subject: Re: A peculiar sum
Newsgroups: sci.math.research

I suppose you already went through this, but I can get it as far as
a product over all primes of a certain number-theoretic infinite series.

In article <31b7cc$71834.301@www.qed.com> you write:
>
>Let phi be the Euler phi-function.
>
>Let sigma be the sum of divisors function.

So if n = Prod(p^(e_p)), phi(n)sigma(n) = Prod(p^(2e_p) - p^(e_p-1)),
the product taken over all primes with  e_p > 0.

>It is trivial to show that  phi(n) sigma(n) is bounded from above
>by n^2. In fact, the largest phi(n)sigma(n) can be is n^2-1 and only
>when n is prime. With a little more work one can show that phi(n)sigma(n)
>is also bounded from below by constant*n^2 and hence:
>
>sum(n=1 to infinity of 1/(phi(n)sigma(n)) converges.

So this is Prod( 1 +  sum e=1 to infinity 1/(p^(2e)-p^(e-1)) ),
the product taken over all primes, i.e., it's
	Prod_p( 1 + p^2 F(1/p) )
where  F(x) = sum( e=1 to infinity of  x^(2(e+1))/(1-x^(e+1)) ).
I would expand and re-index  F  to get
	F(x) = sum( f=2 to infinity, g=2 to infinity   x^(fg) )
	     = sum( n=1 to infinity  d(n) x^n ) + x(x+1)/(x-1)
where  d(n) = number of divisors of  n. Of course this is similar to
the expansion of  zeta(s)^2, but we need a power series  (x^n, not n^s).

>Query: Can this sum be expressed in terms of known constants?

I don't know if  sum( d(n)/p^n ) is the kind of known constant you're
looking for. I would guess not.

dave