Calculate $\sum\limits_{m=0}^{\infty}\sum\limits_{i=0}^{m}\frac{1}{m!} \binom{m}{i} a^{i}b^{m-i} $

Question

Prove that $$\sum\limits_{m=0}^{\infty}\sum_{i=0}^{m}\frac{1}{m!} \binom{m}{i} a^{i}b^{m-i}. =\sum\limits_{i=0}^{\infty}\sum_{m=i}^{\infty}\frac{1}{i!(m-i)!}a^{i}b^{m-i} $$

I'm stacked how changed two sums

I don't know where to start. Any clues?

Answer 1

Think of $(a+b)^m $ and of $e^{a+b}$.

Answer 2

Hint: You can take $\frac{1}{m!}$ outside of your inside sum to get

$$\sum_{m=0}^{\infty} \frac{1}{m!}\left[\sum_{i=0}^m \binom{m}{i} a^ib^{m-i}\right].$$

The inside bracketed sum should look familiar (where else have you seen binomial coefficients in sums?)...

Answer 3

The first sum is indexed by the set $\{ (m,i) \in \mathbb{N} \times \mathbb{N} \mid i \le m \}$, and the second sum is indexed by the set $\{ (i,m) \in \mathbb{N} \times \mathbb{N} \mid m \ge i \}$. The function $(m,i) \mapsto (i,m)$ defines a bijection between these two sets.

Moreover, the quantities $\displaystyle\dfrac{1}{m!} \binom{m}{i} a^ib^{m-i}$ and $\displaystyle\dfrac{1}{i!(m-i)!} a^ib^{m-i}$ are equal.

So in particular the sums are both equal, since they sum the same quantities over the same range of variables.

The first sum can then be evaluated using the binomial theorem and the power series definition of the exponential function.