Why is $\sum^\infty_{n=0}\sum^n_{k=0}\frac{b^k}{k!}\cdot\frac{a^{n-k}}{(n-k)!}=\sum^\infty_{n=0}\frac{(a+b)^n}{n!}$?

Question

The tutor of my algorithms class used the following equation today:

$\sum\limits^\infty_{n=0} \sum\limits^n_{k=0} \frac {b^k} {k!} \cdot \frac {a^{n-k}} {(n-k)!}$ = $\sum\limits^\infty_{n=0} \frac {{(a+b)}^n} {n!}$

When I asked him why we are allowed to use this he just told me that you usually prove it in one of your first maths lectures. However, I am still too stupid to see why this should be true. A hint would be really nice.

I don't know if that is misleading but after having some thoughts about the equation I figured that the part $\sum\limits^n_{k=0} \frac {b^k} {k!} \cdot \frac {a^{n-k}} {(n-k)!}$ is a Cauchy-Product. After finding that out I wrote the first terms of the sum on a piece of paper and always saw that they look pretty similar to the binomial theorem. But I can't find a way to do it for an infinite amount of terms. Just for the given ones I wrote down.

Answer 1

The Binomial theorem states that $$\sum_{k=0}^n{n\choose k}b^k a^{n-k}=(a+b)^n,$$ so $$\sum\limits^\infty_{n=0} \sum\limits^n_{k=0} \frac {b^k} {k!} \cdot \frac {a^{n-k}} {(n-k)!}=\sum\limits^\infty_{n=0} \frac{1}{n!}\left(\sum\limits^n_{k=0} \frac{n!}{k!(n-k)!}{b^k}a^{n-k}\right)\\\quad\quad\quad\quad=\sum\limits^\infty_{n=0} \frac{1}{n!}\sum\limits^n_{k=0} {n\choose k}b^ka^{n-k}\\\quad= \sum\limits^\infty_{n=0} \frac {{(a+b)}^n} {n!}.$$

Answer 2

If you bear in mind that

$$e^x=\sum_{n=0}^\infty{x^n\over n!}$$

the identity we're contemplating is just the series form of

$$e^{a+b}=e^a\cdot e^b$$

Now with the binomial theorem keeping in mind ${n \choose k}=n!/k!(n-k)!$

$$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}$$

We can prove it directly