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In 1737, Euler discovered that if $ f(n) $ is multiplicative and $ \sum f(n)/n^{s} $ converges absolutely for ${\rm Re}(s) > \sigma_a$ then we have \begin{equation} \sum_{n=1}^{\infty} \frac{f(n)}{n^s} ~=~ \prod_p \Bigg\{ 1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+ \cdots \Bigg\} \end{equation} and, especially, if $f$ is completely multiplicative we have \begin{equation} \sum_{n=1}^{\infty} \frac{f(n)}{n^s} ~=~ \prod_p \frac{1}{1-f(p)/p^{s}}~~~~~{\rm if} ~~{\rm Re}(s)>\sigma_a. \end{equation}

I found an example in Wikipedia (https://en.wikipedia.org/wiki/Euler_product) like

\begin{equation} \frac{\pi}{4}~=~ \sum_{n=1}^{\infty} \frac{f(n)}{n},~~~~~~~~{\rm where}~~~~f(n)=\begin{cases}(-1)^{(n-1)/2} & {\rm if} \ n \ {\rm odd}, \\ 0 & {\rm if } \ n \ {\rm even}, \end{cases} \end{equation} so the theorem gives \begin{equation} \frac{\pi}{4}~=~ \prod_{p \not= 2} \frac{1}{1-f(p)/p} ~=~ \prod_{p\not=2} \frac{p}{p-(-1)^{(p-1)/2}}~=~\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot \frac{13}{12}\cdots. \end{equation} However, this example does not converge absolutely but conditionally. If this example holds, how can I prove it though it converges conditionally? Is there any other additional condition needed or should I apply a different method?

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    $\begingroup$ Apply the method for $s>1$ and let $s\to 1$ should work, though this will need some justification. (I don't really like the way you wrote the final formula, it almost looks as if two infinite products were taken.) $\endgroup$
    – Christian Remling
    Commented Dec 11, 2016 at 6:10
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    $\begingroup$ @ChristianRemling that is more subtle for products than sums. There is not a simple version of Abel's theorem for infinite products. See Examples 3.5 and 5.13 in math.uconn.edu/~kconrad/articles/eulerprod.pdf. $\endgroup$
    – KConrad
    Commented Dec 11, 2016 at 14:01
  • $\begingroup$ @ChristianRemling I cleaned up the last formula so it is not written like divergent products anymore. $\endgroup$
    – KConrad
    Commented Dec 11, 2016 at 14:16
  • $\begingroup$ Two related MO questions that you might enjoy and benefit from: mathoverflow.net/questions/63714/… and mathoverflow.net/questions/63787/… $\endgroup$
    – GH from MO
    Commented Dec 11, 2016 at 14:43

1 Answer 1

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You are right to question this. The product $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ (where $\chi = (-1/\cdot)$ is the Dirichlet character mod $4$) does converge, and the limit is $L(1,\chi) = \pi/4$ as expected; But this requires justification $-$ indeed it is equivalent to the non-vanishing of the Dirichlet function $L(s,\chi)$ on the edge $s = 1+it$ of the critical strip, which is also what you need to prove the analogue of the Prime Number Theorem for primes in arithmetic progressions mod $4$. (Taking logarithms, we see that $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ converges if and only if $\sum_p \chi(p)/p$ converges, since this sum differs from the product's logarithm by an absolutely convergent sum $\sum_p O(1/p^2)$; getting from $\sum_p \chi(p)/p$ to $L(s,\chi)$, and then showing that the product $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ actually converges to $L(1,\chi)$, is a classical chapter of analytic number theory.)

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  • $\begingroup$ Yes, but this is not true for every Euler product. The Dirichlet L-functions $L(s,\chi)$ are special because their coefficients are periodic, so they are entire and have a functional equation, from which we know the vertical density of zeros $\beta$ in the critical strip, allowing us to write $\frac{L'(s,\chi)}{L(s,\chi)} = -\sum_\beta \frac{1}{s-\beta}$, from which we get the explicit formula $\sum_{p^k < x} \frac{\chi(p^k)}{k} = \sum_\beta li(x^{\beta})+\mathcal{O}(1)$, $\endgroup$
    – reuns
    Commented Dec 11, 2016 at 18:45
  • $\begingroup$ so that if $L(s,\chi)$ has no zeros for $Re(s) >\sigma$ then $\sum_{p^k < x} \frac{\chi(p^k)}{k} = \mathcal{O}(x^{\sigma+\epsilon})$ and $\log L(s,\chi) = s \int_1^\infty (\sum_{p^k < x} \frac{\chi(p^k)}{k}) x^{-s-1}dx$ converges for $Re(s) >\sigma$. $\endgroup$
    – reuns
    Commented Dec 11, 2016 at 18:46
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    $\begingroup$ I didn't claim, nor (I think) implied, that it's true for every Euler product. $\endgroup$
    – Noam D. Elkies
    Commented Dec 11, 2016 at 19:45

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