Problem in Multiplication of Series

Question

given two series:

$$\sum_{n=0}^{\infty}a_{n}=a_{0}+a_{1}+\, ...\, +a_{n}+\,...\,\,\,\,\,\,\,\,(\rm I)$$

$$\sum_{n=0}^{\infty}b_{n}=b_{0}+b_{1}+\, ...\, +b_{n}+\,...\,\,\,\,\,\,\,\,(\rm {II})$$

we form the series

$$\sum_{n=0}^{\infty}c_{n}=\sum_{n=0}^{\infty}(a_{0}b_{n}+a_{1}b_{n-1}+...+a_{n-1}b_{1}+a_{n}b_{0})\,\,\,\,\,\,\,\,(\rm {III})$$

the product of (I) and (II) in Cauchy's form.

Even if (I) and (II) converge and have ordinary sum A and B , the series (III) may fail to converge. How ever,we still have the following :

THEOREM : If the series (I) and (II) converge and have ordinary sums A and B , then the series (III) is summable by Poisson-Abel method to the generalized sum AB.

this theorem just proved implies Abel's theorem on the multiplication of series.

PROBLEM:

prove that :

$$\sum_{m=0}^{n}\frac{(2m-1)\,.\,(2m-3)\,...\,3\,.\,1}{2m\,. (2m-2)\,...\,4\,.\,2}\times\frac{(2n-2m-1)\,.\,(2n-2m-3)\,...\,3\,.\,1}{(2n-2m)\,.\,(2n-2m-2)\,...\,4\,.\,2}=1$$

where $(2m-1)\,.\,(2m-3)\,...\,3\,.\,1 $ and $2m\,. (2m-2)\,...\,4\,.\,2$ are formally equal to $1$ if $m=0$

Answer 1

We have, using the properties of the double factorial, that $$S\left(n\right)=\sum_{m=0}^{n}\frac{\left(2m-1\right)!!}{\left(2m\right)!!}\frac{\left(2n-2m-1\right)!!}{\left(2n-2m\right)!!}=\sum_{m=0}^{n}\dbinom{2m}{m}\frac{1}{4^{m}}\dbinom{2\left(n-m\right)}{n-m}\frac{1}{4^{n-m}} $$ $$=\frac{1}{4^{n}}\sum_{m=0}^{n}\dbinom{2m}{m}\dbinom{2\left(n-m\right)}{n-m}. $$ Now note that from the generalized binomial theorem we have $$\sum_{n=0}^{\infty}\dbinom{2n}{n}x^{n}=\frac{1}{\sqrt{1-4x}} $$ hence from the Cauchy product we have$$\sum_{n=0}^{\infty}\sum_{m=0}^{n}\dbinom{2m}{m}\dbinom{2\left(n-m\right)}{n-m}x^{n}=\left(\sum_{n=0}^{\infty}\dbinom{2n}{n}x^{n}\right)^{2}=\frac{1}{1-4x} $$ but $$\frac{1}{1-4x}=\sum_{n=0}^{\infty}4^{n}x^{n} $$ hence equating the coefficients we have $$\sum_{m=0}^{n}\dbinom{2m}{m}\dbinom{2\left(n-m\right)}{n-m}=4^{n} $$ so $$S\left(n\right)=\color{red}{1}.$$