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

Hint $\sum_\limits{i+j=k}$ doesn't mean $\sum_\limits{i+j=k}^\infty$ as you wrote. What it really means is
$$\sum_\limits{i+j=k} u_{i, j} = \sum_\limits{i=0}^k u_{i, k-i}$$
