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 *p*_{1}, *p*_{2}, …, *p*_{k} are distinct primes, and the *m*_{1}, *m*_{2}, …, *m*_{k} 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

.

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Tags: Geometric Series, Math, Monday Math, Prime Factors, Prime Numbers, Riemann Zeta Function

This entry was posted on February 23, 2009 at 12:05 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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