Hi I want to prove that for all $x,y\in R$, the following holds
$$(\sum_{n=0}^{\infty}\frac{x^n}{n!})(\sum_{n=0}^{\infty}\frac{y^n}{n!})=\sum_{n=0}^{\infty}\frac{(x+y)^n}{n!}$$
without using $e$ nor $\sum_{n=0}^\infty \frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$
What I know:
$$LHS = 1\sum_{n=0}^{\infty}\frac{y^n}{n!}+ x\sum_{n=0}^{\infty}\frac{y^n}{n!} +\frac{x^2}{2!}\sum_{n=0}^{\infty}\frac{y^n}{n!}+\cdots$$
$$=\ \ 1\ \ \ +\ y\ \ \ +\ \frac{y^2}{2!}\ \ \ +\ \frac{y^3}{3!}\ \ \ +\ \cdots\ \ $$
$$\ +\ x\ \ +\ xy\ \ +\ x\frac{y^2}{2!}\ \ +\ x \frac{y^3}{3!}\ \ +\ \ \cdots$$
$$\ +\frac{x^2}{2!}+\frac{x^2}{2!}y+\frac{x^2}{2!}\frac{y^2}{2!}+\frac{x^2}{2!} \frac{y^3}{3!}+\ \cdots$$
$$\vdots\hspace{6cm}$$
And if we see the expression above as a matrix with $a_{1,1}=1, \ a_{1,2}=y, \ a_{2,1}=x,$ etc, it's clear that it equals:
$$= a_{1,1} + (a_{1,2}+a_{2,1}) + (a_{1,3}+a_{2,2}+a_{3,1}) + \cdots$$
$$=1 + (x+y) + \frac{(x+y)^2}{2}+\cdots=RHS\hspace{1.5cm}$$
But this seems really not rigorous, so I was wondering if there's a way to rigorously prove the equation, and if I'm not on the right path, what could be a way to approach it.
Best Answer
First, prove that $\sum_{n\ge 0} \frac {x^n}{n!}$ and $\sum_{n\ge 0} \frac{y^n}{n!}$ converge absolutely by using the Ratio Test and compute their Cauchy Product: $$ \begin{align} \left(\sum_{n\ge 0} \frac {x^n}{n!}\right)\left(\sum_{n\ge 0} \frac {y^n}{n!}\right)&=\sum_{n\ge 0} c_n = \sum_{n\ge 0} \sum_{i=0}^n \frac{x^i}{i!}\frac{y^{n-i}}{(n-i)!} \\ &=\sum_{n\ge 0} \sum_{i=0}^n \frac 1{n!}\binom{n}{i}x^iy^{n-i}=\sum_{n\ge 0}\frac 1{n!}(x+y)^n \end{align} $$ Where we used the Binomial Theorem in the last equality.