Help understanding this double sum in Feynman diagram cancellation rule

Solution 1:

The equation you wrote is valid. To prove it, start with equation for the Cauchy product of two series, $$ \left(\sum_{k=0}^\infty a(k)\right)\left(\sum_{h=0}^\infty b(h)\right)=\sum_{n=0}^\infty \sum_{k=0}^n a(k)b(n-k) $$ which is valid for any functions $a,b:\mathbb \{0,1,2\dots\}\to \mathbb C$ for which $\sum_{k}a(k)$ and $\sum_k b(k)$ converge absolutely. Apply this equation to $$ a(k)=(-1)^kf(k)/k!,\\ b(h)=(-1)^h g(h)/h!, $$ to get $$ \begin{align} \left(\sum_{k=0}^\infty (-1)^k{f(k)\over k!}\right) \left(\sum_{h=0}^\infty (-1)^h {g(h)\over h!}\right) &=\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^kf(k)}{k!}\cdot \frac{(-1)^{n-k}g(n-k)}{(n-k)!} \\&=\sum_{n=0}^\infty \sum_{k=0}^n (-1)^n \frac1{\color{blue}{n!}}\cdot \frac{\color{blue}{n!}}{k!(n-k)!} f(k)g(n-k) \\&=\sum_{n=0}^\infty \sum_{k=0}^n (-1)^n \frac1{{n!}}\cdot \binom{n}k f(k)g(n-k) \end{align} $$ Notice how we needed to multiply and divide by $n!$ in order to form the binomial coefficient $\binom{n}k$.