6
$\begingroup$

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.

  • 2
    Binomial theorem. – 2017-01-17
  • 1
    Rewrite $\frac{1}{k!}\frac{1}{(n-k)!} = \frac{1}{n!}{n\choose k}$ and use the binomial theorem "in reverse" on the left hand side or expand $(a+b)^n = \sum {n\choose k}b^k a^{n-k}$ on the right hand side. btw the title does not match the question! – 2017-01-17
  • 0
    I have edited the title to match the content of the body of the question. If your actual question truly was about proving the identity you claimed in the original title, then please ask a new question in which *that* is in both the title *and* the body. (*side note: thank you for trying to typeset your equations, but you have far too many unnecessary { }'s and spaces in use, making it difficult to repurpose*) – 2017-01-17
  • 0
    @JVMoarvitz Thank you... I am new to this whole Mathjax thing, messed up first and forgot to paste the right one in the topic. – 2017-01-17

2 Answers 2

2

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

0

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

  • 0
    The connection with the property of the exponential looks nice, however it is not straightforward to me how you get $e^ae^b$ (two infinite series) from $\sum\limits^\infty_{n=0} \sum\limits^n_{k=0} \frac {b^k} {k!} \cdot \frac {a^{n-k}} {(n-k)!}$ (one infinite series of finit sums). – 2017-01-17