Monday Math 60

Let ω(n) be the number of distinct prime factors of the integer n, with us defining ω(1)=0 (see, for example, here). What, then, is the value of the series ?
How about ?

Note first that if a and b are mutually prime, then , and thus .
Now the fundamental theorem of arithmetic tells us that every integer n>1 has a unique prime factorization , where the p1, p2, …, pk are distinct primes, and the m1, m2, …, mk are positive integers (note that k=ω(n)). Then, for n>1,  and .
We see then that the prime product
, when expanded into a sum, will contain terms whose numerator is of the form , with corresponding denominator , with each such combination appearing once. But these are just  and  , with each positive integer n appearing once and only once. Thus

and as , and  for k>0, we have

Now, via the geometric series,

so

and thus
.

Now, we showed previously that , and so
.
To find this last product, note that . Thus
, and again using , we see that
, and so
.

For s=2, this then answers our original question:
, and using  and  (see here), we find
.
Similarly, 
(see here), and so
.