In the following post we find an interesting calculation of $$ \sum_{n\ge 1} \frac{\mu(n)^2}{n\varphi(n)}.$$ I had been trying to do this calculation myself, setting the slightly more ambitious goal of finding a closed form of $$ G(s) = \sum_{n\ge 1} \frac{\mu(n)^2/(n\varphi(n))}{n^s}$$ in some half plane. I gave up on that (when I should have pursued the evaluation of the constant) and I think I can show that there is no such closed representation as a finite product/quotient using only terms of the form $\zeta(ks)$ ($k>0$) and $\zeta(s+k)$ ($k$ an integer). This is because the Euler products for these two are $$ \prod_p \left(1-\frac{1}{p^{ks}}\right)^{-1} \quad \text{and} \quad \prod_p \left(1-\frac{1}{p^s} \frac{1}{p^k}\right)^{-1}$$ Putting $z = \frac{1}{p^s}$ for $s$ a variable we see that any such product consists of products of terms of the form $$ \frac{1}{1-z^k} \quad \text{and} \quad \frac{1}{1-\frac{z}{p^k}},$$ i.e. the roots and singularities relative to $z$ of these are on the unit circle or at integer powers of primes. However, the Euler product for $G(s)$ is $$\prod_p \left( 1 + \frac{1}{p(p-1)} \frac{1}{p^s}\right)$$ which in terms of $z$ is $$ 1 + \frac{1}{p(p-1)} z$$ with root $z = -p (p-1)$. Ergo, no such representation of $G(s)$ as a product/quotient of the above Zeta function terms can exist. My question is, is this correct? Or does there exist a closed form of $G(s)$?

For example, the Euler product for $$H_1(s) = \sum_{n\ge 1} \frac{2^{\omega(n)}}{n^s}$$ where $\omega(n)$ is the number of prime divisors, is $$\prod_p \left( 1 + 2\frac{1/p^s}{1-1/p^s}\right) = \prod_p \left( 1 + 2\frac{z}{1-z}\right)$$ which gives $z=-1$, on the unit circle, and hence $H_1(s)$ has a closed form representation in terms of products and quotients of terms of the form $\zeta(ks)$. The root at $z=-1$ predicts a term in $\zeta(2s)$ and the singularity at $z=1$ a term in $\zeta(s).$ Indeed $$ H_1(s) = \frac{\zeta(s)^2}{\zeta(2s)}.$$

Similarly, the Euler product for $$H_2(s) = \sum_{n\ge 1} \frac{\varphi(n)}{n^s}$$ is $$\prod_p \left( 1 + \frac{p-1}{p^s-p}\right) = \prod_p \left( 1 + \frac{p-1}{1/z-p}\right)$$ which gives $z=1$, on the unit circle, and hence $H_2(s)$ has a closed form representation like $H_1(s).$ The root at $z=1$ predicts an inverse of $\zeta(s)$ and the singularity at $1/p$ corresponds to the term in $\zeta(s-1).$ Indeed $$H_2(s) = \frac{\zeta(s-1)}{\zeta(s)}.$$ Note that we discovered this automatically and without knowing about the Dirichlet convolution $$\varphi\star 1 = n.$$

What we have here is an algorithm for the discovery of closed forms of Dirichlet series from their Euler products. Just examine the roots and singularities according to their multiplicities and include the corresponding terms. Any takers? Maple and Mathematica come to mind.

E.g. to find a closed form of $$\prod_p \left(1 + \frac{z}{(1-z)^2}\right)$$ note that it has the two primitive roots of $1-z^6$ as roots. So include $1/\zeta(6s)$ for these roots and multiply by $\zeta(s),\zeta(2s)$ and $\zeta(3s)$ to cancel those that are not primitive. This is a representation of the fact that $${\frac {1-{z}^{6}}{ \left( 1-z \right) \left( 1-{z}^{2} \right) \left( 1-{z}^{3} \right) }} =1+{\frac {z}{ \left( 1-z \right) ^{2}}}.$$

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    $\begingroup$ $\mu(n)^2/n/\varphi(n)$ is ambiguous. $\endgroup$ – joriki Nov 10 '12 at 5:16

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