How can you prove that
$\displaystyle \prod_{n=1}^{\infty}(1-x^{n})^{\mu(n)/n}=e^{-x}$
for $\displaystyle |x|<1$.
Here, $\displaystyle \mu(n)$ is the Mobius function.
Thanks in advance.
How can you prove that
$\displaystyle \prod_{n=1}^{\infty}(1-x^{n})^{\mu(n)/n}=e^{-x}$
for $\displaystyle |x|<1$.
Here, $\displaystyle \mu(n)$ is the Mobius function.
Thanks in advance.
What have you tried? Begin by taking the logarithm of both sides to get that, if $\displaystyle F(x)$ is our function, $\displaystyle \displaystyle \log(F(x))=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\log \left(1-x^n\right)=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{\mu(n) }{n}\frac{x^{mn}}{m}$. But show that the coefficient of $\displaystyle x^n$ in this is $\displaystyle \displaystyle \frac{-1}{n}\sum_{d\mid n}\mu(d)$ which I'm sure you know is $\displaystyle -1$ if $\displaystyle n=1$ and $\displaystyle 0$ otherwise. Thus, $\displaystyle \log(F(x))=-x$.