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A product

Let \phi denote the golden ratio , \mu the Möbius function and \varphi  Euler’s totient function. Prove that:

    \[\prod_{n=1}^{\infty} \left ( 1 - \frac{1}{\phi^n} \right )^{\frac{\mu(n) - \varphi(n)}{n}} = e\]

Solution

We are using the following facts:

(1)   \begin{equation*} \sum_{d \mid m} \varphi(d) = m \end{equation*}

(2)   \begin{equation*} \sum_{d \mid m} \mu(d) = \left\{\begin{matrix} 1 & , & m=1 \\ 0 & , & m>1 \end{matrix}\right. \end{equation*}

Hence,

    \begin{align*} \sum_{n=1}^{\infty} \frac{\mu(n) - \varphi(n)}{n}\log\left(1 - \frac{1}{\varphi^n} \right) &=-\sum_{n=1}^{\infty}\frac{\mu(n) - \varphi(n)}{n} \sum_{k=1}^{\infty} \frac{1}{k\varphi^{kn}} \\ &=\sum_{m=1}^{\infty} \frac{1}{m \varphi^m} \sum_{d|m} \left(\varphi(d) - \mu(d) \right) \\ &= \sum_{m=2}^{\infty} \frac{1}{\varphi^m}\\ &= \frac{1}{\varphi^2} \frac{1}{1- \frac{1}{\varphi}}\\ &= \frac{1}{\varphi^2-\varphi} \\ &= 1 \end{align*}

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