Can anyone help with find the properties associated with $\prod$ of product

Question

I can easily find some commonly used properties of $\displaystyle\sum $ but I can't seem to find anything about commonly known formulas of $\displaystyle\prod$ except $\displaystyle\prod_{i=1}^{n} i = n! $

Thanks! Also do let me know what did you type in the search bar.

Answer 1

One key word is "finite products". Some hints are given in introductory combinatorics books, e.g. in the following books.

Riordan, J.: An Introduction to Combinatorial Analysis. Princeton University Press, 1958

Comtet, L.: Advanced Combinatorics: The Art of Finite and Innite Expansions. Reidel, 1974

Graham, R. L.; Knuth, D. E.; Patashnik, O.: Concrete Mathematics: A Foundation for Computer Science. Addison-Wesley, 1994

Charalambides, Ch. A.: Enumerative combinatorics. CRC Press, 2002

Answer 2

$$ \log\prod ?=\sum\log\,? $$ allows you to transfer properties of summations to products.

Answer 3

Examples: The Wallis Product for $\pi.$ And Euler's product $\sin\pi x=\pi x \prod_{n=1}^{\infty}(1-x^2/n^2).$ Neither of these is easy to prove. See "Wallis product" in Wikipedia.

A useful theorem, which you can find (for example) in the old classic Infinite Sequences And Series, by Bromwich:

(1). If $a_n\geq 0$ for every $n$ then $\sum_{n=1}^{\infty}a_n$ converges iff $\prod_{n=1}^{\infty}(1+a_n)$ converges.

(2). If $1>a_n\geq 0$ for every $n$ then $\sum_{n=1}^{\infty}a_n$ converges iff $\prod_{n=1}^{\infty}(1-a_n)>0.$

Euler used (2) to show that $\sum_{p\in P}\;(1/p)=\infty,$ where $P$ is the set of primes (and an immediate corollary, a proof that $P$ is infinite, obtained by analytic methods), as follows:

For $2\leq M\in \Bbb N$ let $P(M)$ be the set of primes that are not more than $M.$ We have $$\prod_{p\in P}(1-1/p)^{-1}\geq \prod_{p\in P(M)}(1-1/p)^{-1}=$$ $$=\prod_{p\in P(M)}(\sum_{j=0}^{\infty}p^{-j})>$$ $$>\prod_{p\in P(M)}(\sum_{j=0}^Mp^{-j}).$$ Now if the last expression above is completely expanded, every $n^{-1}$ for $1\leq n\leq M$ will appear at least once as a term (Exactly once, actually, because of unique prime decomposition). There will generally be a whole lot of other terms as well. So for every $M\geq 2 $ we have $$\prod_{p\in P}(1-1/p)^{-1}>\sum_{n=1}^Mn^{-1}.$$ But since $\sum_{n=1}^{\infty}n^{-1}$ diverges we must have $\prod_{p\in P}\;(1-1/p)^{-1}=\infty, $ so $\prod_{p\in P}\;(1-1/p)=0.$ By (2), therefore $\sum_{p\in P}\;(1/p) =\infty.$