# How to find a power series representation for a divergent product?

Euler used the identity $$\frac{ \sin(x) }{x} = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2 } \right) = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n + 1)! } x^{2n}$$ to solve the Basel problem. The product is obtained by noting that the sine function is 'just' an infinite polynomial, which can be rewritten as the product of its zeroes. The sum is found by writing down the taylor series expansion of the sine function and dividing by $x$.

Now, I am interested in finding the sum representation of the following product: $$\prod_{n=1}^{\infty} \left(1 - \frac{x}{n \pi} \right) ,$$ which is divergent (see this article).

The infinite sum representation of this product is not as easily found (at least not by me) because it does not have an obvious formal representation like $\frac{\sin(x)}{x}$ above.

Questions: what is the infinite sum representation of the second product I mentioned? How does one obtain this sum? And is there any 'formal' represenation for these formulae (like $\frac{\sin(x)}{x}$ above).

Not sure if there is a precise sense in which this is meaningful, as the product is divergent for all $x\neq 0$, but the function $$f(x):=\frac{1}{\Gamma\left(1-\frac{x}{\pi}\right)}$$ has simple zeroes at precisely the positive multiples of $\pi$, and satisfies $f(0)=1$. The reflection formula for $\Gamma$ shows that $$f(x)f(-x)=\frac{\sin(x)}{x}=\prod_{n=1}^\infty \left(1-\frac{x}{n\pi}\right)\left(1+\frac{x}{n\pi}\right).$$

• Ah thank you. So now the question is: What is the power series expansion of $\frac{1}{ \Gamma (1 - \frac{x}{\pi} )}$ ? Commented Apr 29, 2013 at 14:40
• @MaxMuller The coefficients in the power series are polynomials in $\pi$, $\gamma$, and values of $\zeta(s)$ at positive integers (with rational coefficients). I don't think it has a particularly nice form, though. Commented Apr 29, 2013 at 15:03
• It becomes fully meaningful once you apply the Weierstrass factorization Commented Feb 25, 2021 at 16:41

I never gave the full answer :)

$$\prod_{n=1}^{\infty} (1-x/n)$$ When analysing a product it's often easiest to consider the form $$\prod_{n=1}^{\infty} (1+f(n))$$ given that $$\sum_{n=1}^{\infty}f(n)^m=G(m)$$; $$f(n)=-x/n$$ Then $$G(m)=(-x)^m\zeta(m)$$.

$$\prod_{n=1}^{\infty} (1-x/n)=e^{\sum_{m=1}^{\infty}\frac{(-1)^{m+1}x^m\sum_{n=1}^{\infty}f(n)^m}{m}}$$ Because $$\zeta(1)$$ is the only part of the product that goes to 0 (e.g. e^-infty), the regularized product will tend to 0 just as $$\zeta(1)$$ goes to infinity. However this is easy fixable:

$$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=e^{\sum_{m=2}^{\infty}\frac{(-1)^{m+1}x^m\sum_{n=1}^{\infty}f(n)^m}{m}}=e^{\sum_{m=1}^{\infty}\frac{- x^{m+1}\zeta(m+1)}{m+1}}$$ which does converge. To show the above more well known representation,

$$\sum_{n=1}^{\infty}-\frac{x^{n+1}\zeta(n+1)}{n+1}=\int_{0}^{-x}\sum_{n=1}^{\infty}(-1)^nz^n\zeta(n+1)dz=\int_{0}^{-x}-H_zdz=-\ln((-x)!)+x\gamma$$ $$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=\frac{e^ {x\gamma}}{(-x)!}$$ And now with the answer I posted 6 years ago, and using the refined stirling numbers: $$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=\frac{e^ {x\gamma}}{(-x)!}=\bigg(1-x^2\zeta(2)/2-x^3 2\zeta(3)/6 +x^4 \big(3\zeta(2)^2-6\zeta(4)\big)/24+... \bigg)$$

So we can also say with p goes to inf: $$p^x\prod_{n=1}^{p} (1-x/n)=\frac{1}{(-x)!}=$$

We can rewrite our G(m) with $$G(1)=-x\gamma$$ and use this in the formula given above to write a polynomal representation, which is easier then multiply our previous polynome by $$e^{-x\gamma}$$.

$$p^x\prod_{n=1}^{p} (1-x/n)= \bigg(1+-x\gamma+x^2 \frac{\gamma^2-\zeta(2)}{2}+x^3 \frac{-\gamma^3+3\gamma\zeta(2)-2\zeta(3)}{3!}+...\big)=\frac{1}{(-x)!}$$

I find it hard to clearly explain these refined stirling numbers but if you are throwing in a bit of time, consider the following ways to represent the found product representation. It's going to be vague but i clarify it if you want.

The "nice" form is in knowing g(m) for all m. There are a lot of way to represent the refined stirling numbers, especially within this context. Lots of it is related to partitions and ways to "write" a number. e.g. for the x^4 term, $$4=1+1+1+1$$ so we get $$g(1)^4*4!/4!/1^4$$ and the next term is $$1+1+2; G(1)^2 G(2) * 4!/1^2/2^1/2$$ as coefficient. So we divide by the a*s^a if you have a G(s)^a. another more intunitive way is to see it as the "unique" combination of all outcomes. Another way is to write it as these unique combination, but use sum of sums (particular cool! And very easy to image things). You can also achieve it algebraric with writing the e powers out, but that's a real hassle. And if you want another way to represent these refined stirling numbers, you can construct them by using previous found stirling numbers and binominals, which is the most efficient.

$$\prod_{i=b}^c 1+a_i$$= $$1+\sum_{i=b}^{c} (a_i)+$$ $$1/2!((\sum_{i=b}^c (a_i))^2-\sum_{i=b}^c (a_i)^2$$ $$1/3!((\sum_{i=b}^c (a_i))^3-3\sum_{i=b}^c (a_i)^2\sum_{i=b}^c (a_i)+2\sum_{i=b}^c (a_i)^3)$$ $$1/4!((\sum_{i=b}^c (a_i))^4-6(\sum_{i=b}^c (a_i))^2\sum_{i=b}^c (a_i)^2+3(\sum_{i=b}^c (a_i)^2)^2+8(\sum_{i=b}^c (a_i)^3)\sum_{i=b}^c (a_i)-6\sum_{i=b}^c (a_i)^4)$$