[Guest]

Saw this a l o n g time ago in a math book on number theory..

Let,

S(p) = 1 + 2^p + 3^p + 4^p + 5^p + ... oo (series over all natural numbers)


F(x, p) = 1 + x^p + x^(2p) + x^(3p) + ... oo

G(p) = F(2, p) * F(3, p) * F(5, p) * F(7, p) * ... oo (series over all primes)

Prove that,

S(p) = G(p)

assuming the series converges for a given p.

[Guest]

I saw only positive numbers, so wouldn't all equal infintity - thus they are equal? It's been a LONG time since I've done functions. So I'm looking for an easy out.

[Guest]

I forgot to add my comment into a Spoiler

Edited August 17, 2009 by jerbil

[Guest]

This fundamental result, first noted by Euler, marks the first observation made in most text books on the Riemmannian Zeta Function.

Consider the n-th prime number p. Then 1/(1 - 1/p^s) = 1 + 1/p^s + 1/(p^2)^s + 1/(p^3)^s + ..... . This is the crucial manouevre.

Every integer n may be parsed into a product of the powers of a set of primes. For example 23 is the product of 23 to the first power and all of the other primes to the zero-th power. 180 is 2^2 * 3^2 * 5^1 multiplied by all the other primes to the zero-th power.

Now one looks at the infinite product 1/(1 - 1/2^s) * 1/(1 - 1/3^s) * 1/(1 - 1/5^s) ..... .

When one rewrites this in terms of the expansions described above, one finds that 1/23^s occurs in the second term in the expansion of 1/(1 - 1/23^s) multiplied by the first terms in the expansions relating to all the other primes, 1 in each case.

1/180^s occurs in the third term in the expansion of 1/(1 - 1/2^s) multiplied by the third term in the expansion of 1/(1 - 1/3^s) multiplied by the second term in the expansion of 1/(1 - 1/5^s) multiplied by the first term in the expansions 1/(1 - 1/p^s) for all other primes.

I think that it is now clear that the product for all primes of 1/(1 - 1/p^s) is the same as the sum for all positive integers of the numbers 1/n^s..

Edited August 17, 2009 by jerbil

[Guest]

I should have emphasized that, when I stated that every integer could be parsed into a product of a set of primes, I should have inserted the word "uniquely", as was stated by Euclid.

Likewise, that word should have been inserted in the description regarding the product formula involving primes.

s must be larger than 1, since the sum of the harmonic series 1 + 1/2 + 1/3 + 1/4 + .... diverges. This produces a deceptively simple proof that the number of primes must be infinite.

Suppose that the number of primes is finite. Then even for s = 1 the product formula would produce a finite result, in disagreement with the sum formula. Therefore the number of primes is infinite.

[Guest]

...this is an amazing equality with a proof that is mathematically simple enough to be understood by 5th grade math. thanks for your comments as well! Never saw the equation in that light.

psycho, "p" is negative number smaller than -1. Two infinities are not equal to each other.. Unless you are talking about the cardinality of those..