Partition generation

Calculus Level 4

Let the partition function P ( n ) P(n) enumerate the ways n n can be expressed as a distinct sum of positive integers, e.g. P ( 4 ) = 5 P(4) = 5 since 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1 are the only ways to represent 4 4 .

p prime [ n = 0 P ( n ) p n ] \prod_{p \ \text{prime}} \left[ \sum_{n=0}^{\infty} P(n)p^{-n} \right]

Does the above product converge?

Yes No Undecidable in ZFC

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1 solution

Jake Lai
Aug 17, 2015

It is known that n = 0 P ( n ) x n = m = 1 ( 1 x m ) 1 \displaystyle \sum_{n=0}^{\infty} P(n)x^{n} = \prod_{m=1}^{\infty} (1-x^{m})^{-1} . Converting the sum in question to a product, we get

p prime m = 1 ( 1 p m ) 1 = m = 1 ζ ( m ) \prod_{p \ \text{prime}} \prod_{m=1}^{\infty} (1-p^{-m})^{-1} = \prod_{m=1}^{\infty} \zeta(m)

Since lim s ζ ( s ) = 1 0 \displaystyle \lim_{s \to \infty} \zeta(s) = 1 \neq 0 and ζ ( 1 ) \zeta(1) is divergent, it must then be that the product diverges.

Argh... I thought it said "Does the above product diverge?"

Julian Poon - 5 years, 10 months ago

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