# Functions defined as infinite products

Are there standard references on infinite products of rational functions and their convergence properties? I'd appreciate information on finite products too!

The original motivation for this is the (finite) product $f(n)=\prod_{i=1}^{n-1}(1-\frac{i}{i^2+n})$ that I had to bound some time ago. Applying some calculus (logarithm to convert into sum, relate to a series, bound with an integral) I could show that $f(n)>\frac{1}{\sqrt{n}}-\frac\pi{8n}$ if $n>9$ (and actually $f(n)\sim \frac{1}{\sqrt{n}}-\frac\pi{8n}$ for big $n$) but I was left with the question whether there is a closed form for "my" finite product, or for the corresponding infinite product $\phi(n)=\prod_{i\ge 1}(1-\frac{i}{i^2+n})$(Any information on it would also make my day).

EDIT May 17: the infinite product $\phi(n)$ is zero (see Robert Israel's answer). Nevertheless, the square of $f(n)$ is $(1/n)\prod_{i=1}^{n-1}(1-(\frac{i}{i^2+n})^2)$, and it is still possible that the infinite product $\prod_{i\ge1}(1-(\frac{i}{i^2+n})^2)$ converges.

So, is there a place to look for techniques to deal with such products if the need arises?

• "Infinite product" is a standard topic in complex analysis textbooks. Closed form for your finite product seems unlikely. – Gerald Edgar May 16 '11 at 17:39
• @quim: the function $\phi(n)$ which you define can be handled using results from complex analysis that are often encountered when discussing infinite Blaschke products - I think there is something on this in Rudin's Real & Complex Analysis but I don't have my copy to hand. (Complex analysis is overkill, but because similar things arise when looking at Blaschke products, perhaps those books are more likely to mention the results you need.) – Yemon Choi May 16 '11 at 20:25
• If memory serves correctly, a classic old book: Bromwich, Infinite Series, has a lot about products as well as sums [but disclaimer: I haven't even looked at this book for at least 15 years!] You might have some luck with various English mathematics books written before 1950, since this kind of thing was a lot more in fashion back then, at least in England [presumably due to G.H.Hardy's influence, although I am no historian]. – Zen Harper May 17 '11 at 6:23
• What is written in the edit follows because f(n) equals 1/n times $\prod_{i=1}^{n-1}(1+\frac{i}{i^2+n})$ (playing around with the fractions). – quim May 17 '11 at 8:23
• Yes, $\prod_{i=1}^\infty (1 - (i/(i^2+n))^2 )$ converges (because $\sum_{i=1}^\infty (i/(i^2+n))^2$ converges), and Maple says it is ${\frac { \left( \sin \left( 1/2\,\pi \, \left( 1+\sqrt {1-4\,n} \right) \right) \right) ^{2}}{ \left( \sinh \left( \pi \,\sqrt {n} \right) \right) ^{2}}}$ – Robert Israel May 17 '11 at 18:47

According to Maple, the finite product $f(n) = \frac{\Gamma(n - (1 + \sqrt{1-4n})/2) \Gamma(n-(1-\sqrt{1-4n})/2) \Gamma(1-\sqrt{-n}) \Gamma(1+\sqrt{-n})} {\Gamma(n-\sqrt{-n}) \Gamma(n+\sqrt{-n}) \Gamma((1-\sqrt{1-4n})/2) \Gamma((1+\sqrt{1-4n})/2)}$. The infinite product is 0, because $\sum_i i/(i^2+n)$ diverges.
• Well, as usual with such things, it may have indeterminate form $\infty/\infty$ at isolated values of $n$, but it extends to a continuous function of $n$. – Gerald Edgar May 17 '11 at 15:14
• Actually, the poles of $\Gamma$ being at nonpositive integers, these problems won't arise for positive $n$. For example, $f(n)$ has a singularity (a pole, I think) at $n=0$. But of course that doesn't mean anything for the product. – Robert Israel May 17 '11 at 18:04
For the Maple computation: write $$1 - \frac{i}{i^{2} + n} = \frac{(i - \frac{1}{2} - \frac{\sqrt{1 - 4 n}}{2}) (i - \frac{1}{2} + \frac{\sqrt{1 - 4 n}}{2})}{(i + \sqrt{-n}) (i - \sqrt{-n})}$$ then write your product using four products of the form $$\prod_{i=1}^{n-1} (i + a) = \frac{\Gamma (a+n)}{\Gamma (a+1)}$$
• Great! I got some valuable references to literature and even a closed form for that particular finite product. So thanks a lot. I wonder whether the relationship with $\sqrt{n}$ and $\pi$ is any easier using the Gammas... I'll have to work it out! – quim May 17 '11 at 14:04