Prove or refute $\prod_{k=1}^\infty\left(\sum_{n=0}^\infty p(n)x^{kn}\right)^{\frac{\mu(k)}{k}}=e^{\frac{x}{1-x}}$

Question

I believe that it is possible to combine the identity $(16)$ from this MathWorld related to the Möbius function, with the the generating function for the partition function $p(n)$, see in this Wikipedia to get $$\prod_{k=1}^\infty\left(\sum_{n=0}^\infty p(n)x^{kn}\right)^{\frac{\mu(k)}{k}}=e^{\frac{x}{1-x}},$$ for $0<x<1$.

I would like to know if my calculations and manipulations were rights, then I can do a comparison between your calculations and reasonings if you provide me those and my calculaitons.

Question. Prove or refute previous identity.

Answer 1

Here is a detailed proof based on the ideas in my comments (but without using the $\phi$ function).

We shall work in the ring $\mathbb{Q}\left[ \left[ x\right] \right] $ of power series in one indeterminate $x$ over the field $\mathbb{Q}$.

We shall use the notation $\mathbb{N}$ for $\left\{ 0,1,2,\ldots\right\} $, and the notation $\mathbb{N}_{+}$ for $\left\{ 1,2,3,\ldots\right\} $.

Definition. For each $n\in\mathbb{N}$, let $p\left( n\right) $ be the number of partitions of the number $n$.

Theorem 1. We have $\sum\limits_{n=0}^{\infty}p\left( n\right) x^{n}=\prod\limits_{k=1}^{\infty}\dfrac{1}{1-x^{k}}$.

Theorem 1 is extremely well-known; e.g., it is the first statement in the "Generating function" section of the Wikipedia page about partitions. I am sure you know it; just writing this for reference's sake.

On the other hand, let us introduce the Möbius function:

Definition. Let $\mu:\mathbb{N}_{+}\rightarrow\mathbb{Z}$ be the map that sends every $n\in\mathbb{N}_{+}$ to $ \begin{cases} \left( -1\right) ^{\omega\left( n\right) }, & \text{if }n\text{ is squarefree;}\\ 0, & \text{otherwise} \end{cases} $. Here, $\omega\left( n\right) $ denotes the number of all distinct primes dividing $n$.

The function $\mu$ is called the Möbius function.

Now, your claim is the following:

Theorem 2. We have $\prod\limits_{k=1}^{\infty}\left( \sum\limits_{n=0}^{\infty}p\left( n\right) x^{kn}\right) ^{\mu\left( k\right) /k} =\exp\dfrac{x}{1-x}$.

We are going to prepare for the proof of this theorem by introducing a few more notations:

Definition. Let $n\in\mathbb{N}_{+}$. Then, the summation sign $\sum\limits_{d\mid n}$ shall always mean a sum over all positive divisors $d$ of $n$ (rather than a sum over all divisors $d$ of $n$). For example, $\sum\limits_{d\mid2}d$ is $1+2=3$ (not $1+2+\left( -1\right) +\left( -2\right) =0$).

Definition. Let $\mathcal{A}$ be a logical statement. Then, an element $\left[ \mathcal{A}\right] \in\left\{ 0,1\right\} $ is defined as follows: We set $\left[ \mathcal{A}\right] = \begin{cases} 1, & \text{if }\mathcal{A}\text{ is true};\\ 0, & \text{if }\mathcal{A}\text{ is false} \end{cases} $. This element $\left[ \mathcal{A}\right] $ is called the truth value of $\mathcal{A}$. (For example, $\left[ 1+1=2\right] =1$ and $\left[ 1+1=3\right] =0$.) The notation $\left[ \mathcal{A}\right] $ for the truth value of $\mathcal{A}$ is known as the Iverson bracket notation.

Now, a well-known fact (see, e.g., Proposition 2.4.7 in my notes on Floor and arithmetic functions) says the following:

Theorem 3. Let $n\in\mathbb{N}_{+}$. Then, $\sum\limits_{d\mid n} \mu\left( d\right) =\left[ n=1\right] $.

(There are many alternative notations for $\left[ n=1\right] $; for instance, it can be called $\varepsilon\left( n\right) $ or $\delta_{n,1}$.)

Another well-known fact is the following expansion of the logarithm as a power series (often called the Mercator series):

Theorem 4. We have $\log\left( 1-x\right) =-\sum\limits_{s\geq1} \dfrac{x^{s}}{s}$.

(Here and in the following, $\log$ denotes the natural logarithm.)

Now that we are done with the boilerplate, let us restate Theorem 3:

Proposition 5. Let $n\in\mathbb{N}_{+}$. Then, $\sum\limits_{d\geq1} \sum\limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) =\left[ n=1\right] $.

Proof of Proposition 5. Let $d\geq1$. Then, the sum $\sum \limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) $ can be simplified as follows, depending on whether $d\nmid n$ or $d\mid n$:

• If $d\nmid n$, then this sum $\sum\limits_{\substack{s\geq1;\\ds=n} }\mu\left( d\right) $ has no addends (since there exists no $s\geq1$ satisfying $ds=n$), and thus simplifies to $0$.

• If $d\mid n$, then this sum $\sum\limits_{\substack{s\geq1;\\ds=n} }\mu\left( d\right) $ has exactly one addend (namely, the addend for $s=n/d$), and this addend is $\mu\left( d\right) $. Thus, if $d\mid n$, then this sum $\sum\limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) $ simplifies to $\mu\left( d\right) $.

Combining these two observations, we obtain

(1) $\sum\limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) = \begin{cases} 0, & \text{if }d\nmid n;\\ \mu\left( d\right) , & \text{if }d\mid n \end{cases} $.

Now, forget that we fixed $d$. We thus have proven (1) for each $d\geq1$. Summing the equality (1) up over all $d\geq1$, we obtain

$\sum\limits_{d\geq1}\sum\limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) =\sum\limits_{d\geq1} \begin{cases} 0, & \text{if }d\nmid n;\\ \mu\left( d\right) , & \text{if }d\mid n \end{cases} $

$=\underbrace{\sum\limits_{\substack{d\geq1;\\d\mid n}}}_{=\sum\limits_{d\mid n}}\underbrace{ \begin{cases} 0, & \text{if }d\nmid n;\\ \mu\left( d\right) , & \text{if }d\mid n \end{cases} }_{\substack{=\mu\left( d\right) \\\text{(since }d\mid n\text{)}} }+\sum\limits_{\substack{d\geq1;\\d\nmid n}}\underbrace{ \begin{cases} 0, & \text{if }d\nmid n;\\ \mu\left( d\right) , & \text{if }d\mid n \end{cases} }_{\substack{=0\\\text{(since }d\nmid n\text{)}}}$

$=\sum\limits_{d\mid n}\mu\left( d\right) +\underbrace{\sum \limits_{\substack{d\geq1;\\d\nmid n}}0}_{=0}=\sum\limits_{d\mid n}\mu\left( d\right) =\left[ n=1\right] $ (by Theorem 3).

This proves Proposition 5.

Now, we can prove Theorem 2:

Proof of Theorem 2. For each integer $d\geq1$, we have

$\sum\limits_{n=0}^{\infty}p\left( n\right) \underbrace{x^{dn}}_{=\left( x^{d}\right) ^{n}}=\sum\limits_{n=0}^{\infty}p\left( n\right) \left( x^{d}\right) ^{n}=\prod\limits_{k=1}^{\infty}\dfrac{1}{1-\left( x^{d}\right) ^{k}}$

(this follows from Theorem 1 by substituting $x^{d}$ for $x$)

$=\prod\limits_{k=1}^{\infty}\dfrac{1}{1-x^{dk}}=\prod\limits_{w=1}^{\infty }\dfrac{1}{1-x^{dw}}$

(here, we have renamed the index $k$ as $w$ in the product). Taking logarithms on both sides of this equality, we obtain

$\log\left( \sum\limits_{n=0}^{\infty}p\left( n\right) x^{dn}\right) =\log\left( \prod\limits_{w=1}^{\infty}\dfrac{1}{1-x^{dw}}\right) $

$=\sum\limits_{w=1}^{\infty} \left(-\underbrace{\log\left( 1-x^{dw}\right) }_{\substack{=-\sum\limits_{s\geq1}\dfrac{\left( x^{dw}\right) ^{s}} {s}\\\text{(this follows from Theorem 4 by substituting }x^{dw}\text{ for }x\text{)}}}\right)$

(since the logarithm takes products into sums, and takes quotients into differences)

$=\sum\limits_{w=1}^{\infty}-\left( -\sum\limits_{s\geq1}\dfrac{\left( x^{dw}\right) ^{s}}{s}\right) =\underbrace{\sum\limits_{w=1}^{\infty} }_{=\sum\limits_{w\geq1}}\sum\limits_{s\geq1}\underbrace{\dfrac{\left( x^{dw}\right) ^{s}}{s}}_{=\dfrac{1}{s}x^{dws}}$

(4) $=\sum\limits_{w\geq1}\sum\limits_{s\geq1}\dfrac{1}{s}x^{dws}$.

Since the logarithm takes products into sums and takes powers into multiples, we have

$\log\left( \prod\limits_{k=1}^{\infty}\left( \sum\limits_{n=0}^{\infty }p\left( n\right) x^{kn}\right) ^{\mu\left( k\right) /k}\right) $

$=\sum\limits_{k=1}^{\infty}\left( \mu\left( k\right) /k\right) \log\left( \sum\limits_{n=0}^{\infty}p\left( n\right) x^{kn}\right) $

$=\underbrace{\sum\limits_{d=1}^{\infty}}_{=\sum\limits_{d\geq1}}\left( \mu\left( d\right) /d\right) \underbrace{\log\left( \sum\limits_{n=0} ^{\infty}p\left( n\right) x^{dn}\right) }_{\substack{=\sum\limits_{w\geq 1}\sum\limits_{s\geq1}\dfrac{1}{s}x^{dws}\\\text{(by (4))}}}$

(here, we have renamed the summation index $k$ as $d$ in the outer sum)

$=\sum\limits_{d\geq1}\left( \mu\left( d\right) /d\right) \sum \limits_{w\geq1}\sum\limits_{s\geq1}\dfrac{1}{s}x^{dws}$

$=\underbrace{\sum\limits_{d\geq1}\sum\limits_{w\geq1}\sum\limits_{s\geq1} }_{=\sum\limits_{w\geq1}\sum\limits_{s\geq1}\sum\limits_{d\geq1} }\underbrace{\left( \mu\left( d\right) /d\right) \cdot\dfrac{1}{s}} _{=\mu\left( d\right) /\left( ds\right) }\underbrace{x^{dws}}_{=x^{dsw}}$

$=\sum\limits_{w\geq1}\sum\limits_{s\geq1}\underbrace{\sum\limits_{d\geq1} \mu\left( d\right) /\left( ds\right) \cdot x^{dsw}}_{\substack{=\sum \limits_{n\geq1}\sum\limits_{\substack{d\geq1;\\ds=n}}\mu\left( d\right) /\left( ds\right) \cdot x^{dsw}\\\text{(since each }d\geq1\text{ satisfies }ds=n\\\text{for a unique }n\geq1\text{)}}}$

$=\sum\limits_{w\geq1}\underbrace{\sum\limits_{s\geq1}\sum\limits_{n\geq1} \sum\limits_{\substack{d\geq1;\\ds=n}}}_{=\sum\limits_{n\geq1}\sum \limits_{d\geq1}\sum\limits_{\substack{s\geq1;\\ds=n}}}\mu\left( d\right) /\left( \underbrace{ds}_{=n}\right) \cdot\underbrace{x^{dsw}} _{\substack{=x^{nw}\\\text{(since }ds=n\text{)}}}$

$=\sum\limits_{w\geq1}\sum\limits_{n\geq1}\sum\limits_{d\geq1}\sum \limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) /n\cdot x^{nw}$

$=\sum\limits_{w\geq1}\sum\limits_{n\geq1}\dfrac{1}{n}\underbrace{\sum \limits_{d\geq1}\sum\limits_{\substack{s\geq1;\\ds=n}}\mu\left( d\right) }_{\substack{=\left[ n=1\right] \\\text{(by Proposition 5)}}}\cdot x^{nw}$

$=\sum\limits_{w\geq1}\underbrace{\sum\limits_{n\geq1}\dfrac{1}{n}\left[ n=1\right] \cdot x^{nw}}_{=\dfrac{1}{1}\left[ 1=1\right] \cdot x^{1w} +\sum\limits_{n\geq2}\dfrac{1}{n}\left[ n=1\right] \cdot x^{nw}}$

$=\sum\limits_{w\geq1}\left( \underbrace{\dfrac{1}{1}}_{=1} \underbrace{\left[ 1=1\right] }_{=1}\cdot\underbrace{x^{1w}}_{=x^{w}} +\sum\limits_{n\geq2}\dfrac{1}{n}\underbrace{\left[ n=1\right] }_{\substack{=0\\\text{(since }n\neq1\\\text{(since }n\geq2\text{))}}}\cdot x^{nw}\right) $

$=\sum\limits_{w\geq1}\left( x^{w}+\underbrace{\sum\limits_{n\geq2}\dfrac {1}{n}0\cdot x^{nw}}_{=0}\right) =\sum\limits_{w\geq1}x^{w}=\sum \limits_{w\geq0}\underbrace{x^{w+1}}_{=xx^{w}}$

(here, we have substituted $w+1$ for $w$ in the sum)

$=\sum\limits_{w\geq0}xx^{w}=x\underbrace{\sum\limits_{w\geq0}x^{w}} _{=\dfrac{1}{1-x}}=x\cdot\dfrac{1}{1-x}=\dfrac{x}{1-x}$.

Taking the exponential on both sides of this equality, we obtain

$\prod\limits_{k=1}^{\infty}\left( \sum\limits_{n=0}^{\infty}p\left( n\right) x^{kn}\right) ^{\mu\left( k\right) /k}=\exp\dfrac{x}{1-x}$.

Theorem 2 is proven.