show that $\left(\sum_{i=0}^\infty\frac{a^i}{i!}\right)\left(\sum_{j=0}^\infty\frac{b^j}{j!}\right) = \sum_{k=0}^\infty\frac{(a+b)^k}{k!}$

Question

I need to show that if $\sum_{i=0}^\infty \frac{a^i}{i!}$ is absolutely convergent for all $a\in\mathbb{R}$, then $$\left(\sum_{i=0}^\infty\frac{a^i}{i!}\right)\left(\sum_{j=0}^\infty\frac{b^j}{j!}\right) = \sum_{k=0}^\infty\frac{(a+b)^k}{k!}.$$

We see that the Cauchy product of the two series on the left hand side is defined to be $$\sum_{k=0}^\infty \sum_{i+j=k}^\infty \frac{a^i}{i!}\cdot \frac{b^j}{j!}.$$

I'm having trouble showing that $$\sum_{k=0}^\infty\sum_{i+j=k}^\infty \frac{a^i}{i!}\cdot \frac{b^j}{j!} = \sum_{k=0}^\infty\frac{(a+b)^k}{k!}$$

would anyone be able to help? I know that once I get this then since the two series on LHS on the very top are convergent, then so is the Cauchy product. Thus the statement would be proved.

Answer 1

We have that \begin{align*} \left(\sum_{i=0}^\infty\frac{a^i}{i!}\right)\left(\sum_{j=0}^\infty\frac{b^j}{j!}\right)&=\sum_{k=0}^{\infty}\sum_{(i,j)\in S_k}\frac{a^i}{i!}\cdot \frac{b^j}{j!}\\ &=\sum_{k=0}^{\infty}\sum_{(i,j)\in S_k}\frac{1}{(i+j)!} \binom{i+j}{i}a^ib^j\\ &=\sum_{k=0}^{\infty}\frac{1}{k!} \sum_{i=0}^k\binom{k}{i}a^ib^{k-i}=\sum_{k=0}^{\infty}\frac{(a+b)^k}{k!} \end{align*} where $S_k:=\{(i,j): i\geq 0, j\geq 0, i+j=k\}$.

Answer 2

Hint $\sum_\limits{i+j=k}$ doesn't mean $\sum_\limits{i+j=k}^\infty$ as you wrote. What it really means is

$$\sum_\limits{i+j=k} u_{i, j} = \sum_\limits{i=0}^k u_{i, k-i}$$