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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).

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Let's prove the more general:

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

from which yours follow.

Now,

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

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  • $\begingroup$ How does mine follow from this? I don't see where the $(-1)^n$ would fit in... $\endgroup$ Apr 16 '16 at 21:40
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    $\begingroup$ What you want is to prove that $\exp(x)\exp(-x)=1$. Now note that that by what I proved, $\exp(x)\exp(-x)=\exp(x-x)=\exp(0)=1$. $\endgroup$
    – Aloizio Macedo
    Apr 16 '16 at 21:45
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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 $0^0=1$.

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  • $\begingroup$ Would it be possible to prove this without introducing the binomial coefficient as well? I do understand this solution, but the binomial coefficients are only defined after this problem and proving the binomial theorem would be one of the next problems (and therefore I would expect this can be solved without it). $\endgroup$ Apr 17 '16 at 7:18
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    $\begingroup$ @MrTsjolder: I don't think so. The binomial theorem is at least implicitely in any case used to calculate the inner sum. It's also used in the other answer. But, this is an interesting aspect and you could discuss it with your prof. If he provides another insight, I'm interested in it. Regards, $\endgroup$
    – epi163sqrt
    Apr 17 '16 at 7:39
  • $\begingroup$ My professor could not come up with an alternative unfortunately.... $\endgroup$ Apr 23 '16 at 15:07
  • $\begingroup$ @MrTsjolder: So, you can freely use the binomial theorem also for this example. Many thanks for your feedback. $\endgroup$
    – epi163sqrt
    Apr 23 '16 at 15:20

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