Product with Euler’s totient function

Let $\varphi$ be Euler’s totient function and $x \in (0, 1)$ . Prove that

$\prod_{n=1}^{\infty} \left ( 1 - x^n \right )^{\varphi(n)/n} = e^{x/(x-1)}$

Solution

Let $\Pi$ denote the product. Hence,

$\begin{align*} \ln \Pi &= -\sum_{n=1}^{\infty} \frac{\varphi(n)}{n} \ln \left ( 1 - x^n \right ) \\ &= \sum_{n=1}^{\infty} \varphi(n) \sum_{m=1}^{\infty} \frac{x^{nm}}{nm} \\ &= \sum_{j=1}^{\infty} \frac{x^j}{j} \sum_{d \mid j} \varphi(d) \\ &= \sum_{j=1}^{\infty} x^j \\ &= \frac{x}{1-x} \end{align*}$

and the result follows.

Continuous and periodic

Product with Euler’s totient function

They might interest you:

Continuous and periodic

On the centralizer

Integral representation of gamma

An integral inequality

Derivative at 0

A trigonometric limit

Find the function

A trigonometric value

Limit of a multiple integral

Leave a Reply Cancel reply