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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.

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  • $\begingroup$ Your argument is the key behind Cauchy product of two series. There is also the Dirichlet product where the matrix used is bit different. Cauchy product is suited to power series of type $\sum a_nx^n$ and the Dirichlet product is suited to series of the form $\sum a_nn^{-s} $. $\endgroup$ – Paramanand Singh Nov 27 '18 at 5:41
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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.

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