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In QFT we have a cancellation rule for the Feynman diagram, which involves a factorization of a double sum of the form: $$ \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{(-1)^n}{n!} \binom{n}{k} f(k)g(n-k) = \left(\sum_{k=0}^\infty \frac{(-1)^k}{k!} f(k)\right)\left(\sum_{n=0}^\infty \frac{(-1)^n}{n!} g(n)\right) $$ where $f$ and $g$ are some generic functions.

I find this hard to prove as I am missing factors of $\frac{1}{n!(n-k)!}$.

If it's correct, how to prove it? If it's wrong, what would be the correct form?

wooohooo
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    I suspect you have not reparametrised the $(k,n)$ space correctly? – copper.hat Dec 10 '21 at 21:51
  • Isn't this a direct consequence of the Cauchy product for infinite series: https://en.wikipedia.org/wiki/Cauchy_product#Cauchy_product_of_two_infinite_series – Stefan Lafon Dec 11 '21 at 00:23

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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 $\binom{n}k$.

Mike Earnest
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