Applying Cauchy product to formal power series

Question

I am trying to prove that $$ \left(\sum_{n=0}^{\infty} \frac{1}{n!} x^n\right) \cdot \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n\right) = 1$$ where $\cdot$ is the Cauchy product.

I started computing the product which gives either one of $$\left(\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n} \frac{1}{k!} \cdot \frac{(-1)^{n-k}}{(n-k)!} \right) x^n \right) = \left(\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n} \frac{1}{(n-k)!} \cdot \frac{(-1)^{k}}{k!} \right) x^n \right)$$ After some trying around I thought it might be a good idea to split that sum into two parts. In order to do this properly though, I made a distinction between the case where $n$ is odd and the case where $n$ is even

1. when $n$ is odd, I have $$\left(\sum_{n=0}^{\infty} \left(\sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^{k} \cdot (-1)^{n-k}}{(n-k)!k!} \right) x^n \right) = 0$$ because when $k$ is even, $n-k$ is odd and vice versa.
2. when $n$ is even, I get something like $$\left(\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n/2-1} \frac{(-1)^{k} \cdot (-1)^{n-k}}{(n-k)!k!} + \frac{(-1)^\frac{n}{2}}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!} \right) x^n \right)$$ I know that this is zero for $n \geq 2$, but I do not see how this could be proven.

Am I missing something here or are there easier/better ways to prove this? Any help or guidance would be highly appreciated (maybe also on the tags).

Answer 1

It's also convenient to introduce binomial coefficients.

Starting from one of your series expansions we obtain \begin{align*} \sum_{n=0}^\infty&\left(\sum_{k=0}^{n}\frac{1}{k!}\cdot\frac{(-1)^{n-k}}{(n-k)!}\right)x^n\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}\right)\frac{x^n}{n!}\tag{1}\\ &=\sum_{n=0}^\infty\left(1+(-1)\right)^{n}\frac{x^n}{n!}\tag{2}\\ &=1 \end{align*}

Comment:

• In (1) we expand the expression with $\frac{n!}{n!}$

• In (2) we use the binomial theorem together with $0^n=0$ for $n>0$ and $0^0=1$.

Answer 2

Let's prove the more general:

$$\exp(x+y)=\exp(x)\exp(y),$$

from which yours follow.

Now,

$$\exp(x)\exp(y)=(\sum_n \frac{x^n}{n!})(\sum_m \frac{y^m}{m!})=\sum_{n=0}^{\infty}\sum_{i=0}^{n}\frac{x^i}{i!}\frac{y^{n-i}}{(n-i)!}$$ $$=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{i=0}^{n}\frac{n!}{(n-i)!i!}x^iy^{n-i}$$ $$=\sum_{n=0}^{\infty}\frac{1}{n!}(x+y)^n=\exp(x+y).$$