Generation partition

Calculus Level 3

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

$\prod_{p \ \text{prime}} \left[ \sum_{n=0}^{\infty} P(n)p^{-n}(1-\frac{1}{p}) \right]$

Does the above product converge?

No Undecidable in ZFC Yes

1 solution

Julian Poon
Aug 18, 2015

$\prod _{ p{ prime } } \left[ 1-\frac { 1 }{ p } \right] \times \prod _{ p{ prime } } \sum _{ n=0 }^{ \infty } P(n)p^{ -n }=\prod _{ p{ prime } } \left[ 1-\frac { 1 }{ p } \right] \times \prod _{ p{ prime } } \left[ \prod _{ k=1 }^{ \infty }{ \left[ \frac { 1 }{ 1-\frac { 1 }{ { p }^{ k } } } \right] } \right] \\ =\prod _{ p{ prime } } \left[ 1-\frac { 1 }{ p } \right] \times \prod _{ k=1 }^{ \infty }{ \left[ \prod _{ p{ prime } } \left[ \frac { 1 }{ 1-\frac { 1 }{ { p }^{ k } } } \right] \right] } =\prod _{ p{ prime } } \left[ 1-\frac { 1 }{ p } \right] \times \prod _{ k=1 }^{ \infty }{ \left[ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ k } } } \right] }$

Since $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n } } } \times \prod _{ p{ prime } } \left[ 1-\frac { 1 }{ p } \right] =1$

$\prod _{ p{ prime } } \left[ 1-\frac { 1 }{ p } \right] \times \prod _{ k=1 }^{ \infty }{ \left[ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ k } } } \right] } =\prod _{ k=2 }^{ \infty }{ \left[ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ k } } } \right] } = 2.294856591673313794183$

Moderator note:

With infinite sequences, you have to be careful with your manipulation.

Since $\sum \frac{1}{n} = \infty$ , you have to explain how the product yields 1.

How do you know the final product converges? (Hint: $\displaystyle \sum_{n=2}^{\infty} (\zeta(n)-1) = 1$ .)

Jake Lai - 5 years, 10 months ago

Generation partition

1 solution

Moderator note:

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