How to show that $\sum_{m,n>0} \frac{(m+n)!}{m!n!}a^mb^n$ diverges when $|a|+|b|>1$? $a$ and $b$ are considered to be complex numbers. I try to examine the limit of $\frac{(m+n)!}{m!n!}a^mb^n$, but it seems not work.
How to show that $\sum_{m,n>0} \frac{(m+n)!}{m!n!}a^mb^n$ diverges when $|a|+|b|>1$?
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If we want to check for absolute convergence, we have $$\sum_{m,n>0}\frac{(m+n)!}{m!n!}|a|^{m}|b|^{n}=\sum_{s=2}^{\infty}\sum_{m=1}^{s-1}\frac{s!}{m!(s-m)!}|a|^{m}|b|^{s-m}.$$ By the Binomial formula, this gives $$\sum_{s=2}^{\infty}[(|a|+|b|)^{s}-|a|^{s}-|b|^{s}].$$ For this to converge, we need $|a|+|b|<1.$