Give an example of a canonical product \[ f(z)=\prod_{n=1}^{\infty}\left(1-\frac{z}{a_n}\right) \] that has order exactly \(1\).
Solution
Take \[ \boxed{a_n=n(\log(n+1))^2} \] and set \[ f(z)=\prod_{n=1}^{\infty}\left(1-\frac{z}{a_n}\right). \]
1. The series\[ \sum_{n=1}^{\infty}\frac1{a_n} \]converges.
For large \(n\), \(\log(n+1)\) is comparable to \(\log n\), so \[ \frac1{a_n} =\frac1{n(\log(n+1))^2} \asymp\frac1{n(\log n)^2}. \] The latter series converges by the integral test, since \[ \int_2^\infty\frac{dx}{x(\log x)^2}<\infty. \]
2. Consequently the displayed product is a genus-zero canonical productand defines an entire function.
For genus zero the Weierstrass factor is \(E_0(w)=1-w\). Convergence of \(\sum 1/|a_n|\) is exactly the criterion ensuring that \[ \prod_n E_0(z/a_n) =\prod_n\left(1-\frac z{a_n}\right) \] converges uniformly on compact subsets to an entire function with zero set \((a_n)\).
3. The critical exponent of the zero sequence \((a_n)\) is exactly \(1\).
By 1, the series \[ \sum_n|a_n|^{-1} \] converges, so the critical exponent is at most \(1\).
Now fix \(0<s<1\). Since logarithms grow more slowly than every positive power, for all sufficiently large \(n\), \[ (\log(n+1))^{2s}\le n^{(1-s)/2}. \] Hence \[ a_n^s =n^s(\log(n+1))^{2s} \le n^{(1+s)/2}, \] and therefore \[ \frac1{a_n^s}\ge n^{-(1+s)/2}. \] Because \((1+s)/2<1\), the series on the right diverges. Thus \[ \sum_n|a_n|^{-s}=\infty \] for every \(s<1\). The critical exponent is therefore \(1\).
4. The entire function \(f\) has order exactly \(1\).
By Borel’s theorem, the canonical product associated to a zero sequence has order equal to the critical exponent of that sequence. By 3, \[ \rho(f)=\alpha(a_n)=1. \] Thus the genus-zero product above is an explicit example of exact order one.