Dirichlet series and Euler product

Question

For a multiplicative function $f$, show that we have \begin{equation}\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p\left(\sum_{\nu=0}^{\infty}\frac{f(p^{\nu})}{p^{\nu s}}\right).\end{equation}

My solution: Write $n$ as a product of distinct prime powers, i.e. $n=p_1^{\alpha_1}\times\cdots\times p_r^{\alpha_r}$. Then the LHS is the same as \begin{align}\sum_{n=1}^{\infty}\frac{f(n)}{n^s}&=\sum_{n=1}^{\infty}\frac{f(p_1^{\alpha_1}\times\cdots\times p_r^{\alpha_r})}{p_1^{\alpha_1s}\times\cdots\times p_r^{\alpha_rs}}\\&=\sum_{n=1}^{\infty}\left(\frac{f(p_1^{\alpha_1})}{p_1^{\alpha_1s}}\times\cdots\times\frac{f(p_r^{\alpha_r})}{p_r^{\alpha_rs}}\right)\\&=\sum_{n=1}^{\infty}\left(\prod_{i=1}^r\frac{f(p_i^{\alpha_i})}{p_i^{\alpha_is}}\right).\end{align}

This is quite close to what we want, but not quite. We need to justify swapping the sum and product (I assume this is by absolute convergence?). Furthermore the product is taken over primes from $i$ up to $r$ but I believe in the initial problem we want it to be taken over all primes. Can anyone help?

Answer 1

You want to show for any multiplicative function $f$, \begin{equation}\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p\left(\sum_{\nu=0}^{\infty}\frac{f(p^{\nu})}{p^{\nu s}}\right). \end{equation}

Your question about convergence is a good one. As stated, the equality is purely formal, in that it doesn't consider large $f$. What if $f$ is huge? For example, what if $f(p^k) = p^{k^2}$ (then extended multiplicatively)? Then neither side converges for any $s$.

Thus you either need to assume $f$ is small enough for the Dirichlet series to converge somewhere (in which case it actually converges absolutely somewhere, possibly for a slightly larger $s$; you can prove this on your own, or you can look up the topic of the abscissa of convergence and the abscissa of absolute convergence of Dirichlet series), or alternately make the proof for formal power series.

My solution: Write $n$ as a product of distinct prime powers, i.e. $n=p_1^{\alpha_1}\times\cdots\times p_r^{\alpha_r}$. Then the LHS is the same as \begin{align}\sum_{n=1}^{\infty}\frac{f(n)}{n^s}&=\sum_{n=1}^{\infty}\frac{f(p_1^{\alpha_1}\times\cdots\times p_r^{\alpha_r})}{p_1^{\alpha_1s}\times\cdots\times p_r^{\alpha_rs}}\end{align}

You have assumed that every $n$ can be written as the same product $p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. That is, you've assumed that every integer has the same unique prime factorization. Of course, this isn't true.

This is the source of your problem with the product going only up to $r$.