Can we convert infinite sums into infinite products?

Question

Is there any method to convert an infinite sum (of any type) into an infinite product? That is, a method such that the equality $ \sum_{n = 0}^{\infty}a_n = \prod_{n = 0}^{\infty} f(a_n)$

holds.

Answer 1

Yes, to a more general question. Given a sequence of numbers $\,c_1,c_2,\dots,\,$ define $$p_0:=1,\;\; p_1:=p_0+c_1p_0=(1+c_1)p_0,\;\; p_2:=p_1+c_2p_1=(1+c_2)p_1,\dots\;.$$ Now assuming the limit exists, $$\lim_{n->\infty}p_n=\prod_{n=1}^\infty(1+c_n)=1+c_1p_0+c_2p_1+\dots=1+\sum_{n=1}^\infty b_n,\;b_n:=c_n\prod_{k=1}^{n-1}p_k.$$

Now, to reverse the process, given a sequence of numbers $\,a_0, a_1, \dots\,$ and assume that $a_0\ne 0.$ Using the previous method of defining $\,p_0,p_1,\dots\,$ define $$c_1:=\frac{a_1}{a_0 p_0},\;c_2:=\frac{a_2}{a_0 p_1}, \;\dots,\; c_n:=\frac{a_n}{a_0 p_{n-1}},\quad\dots\quad\textrm{and}\quad\sum_{n=0}^\infty a_n = a_0\prod_{n=1}^\infty (1+ c_n).$$ Finally, letting $q_0:=a_0,\; q_n:=1+c_n$ for $n>0,\;$ then $\,\sum_{n=0}^\infty a_n = \prod_{n=0}^\infty q_n.$

The advantage of this method is that it works for finite sums and products also. It is very similar to Euler's method to convert sums to continued fractions.

An example of the method converts the geometric series into an infinite product $$ 1 + x + x^2 + x^3 +\dots = \Big(1+\frac{x^1}{1}\Big) \Big(1+\frac{x^2}{1+x}\Big)\Big(1+\frac{x^3}{1+x+x^2}\Big)\cdots. $$