I'm not sure how exactly to approach this problem. I believe it has something to do with the Laurent Series of the integrand based on the factorial terms in the summation, but I'm not sure exactly how to get there.
2026-04-02 12:30:47.1775133047
How do I show that $\int_\gamma e^{z+1/z}dz = 2\pi i \sum_{n = 0}^\infty \frac{1}{(n!)(n+1)!} $?
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$$\exp(z+1/z)=\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{m=0}^\infty\frac{z^{-m}}{m!} =\sum_{m,n=0}^\infty\frac{z^{n-m}}{n!m!}.$$ If one needed to find the residue at $z=0$ you need to pull out the coefficient of $z^{-1}$, so look at the terms where $m=n+1$.