I am having trouble with proving following equalities
\begin{align} \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{(n+k)!}{n! k!} x^n = \sum_{k=0}^{\infty} (1-x)^{-k-1} \end{align}
How this equalities holds?
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1$\binom{n+k}{n}=(-1)^n\binom{-k-1}{n}$ (see this answer) – robjohn Jul 26 '19 at 15:13
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@robjohn, Oh there is a formula for combinatorics for negative case!, That was my first experience!. I learn binomial theorem for negative exponents because of you! Thanks – phy_math Jul 26 '19 at 15:24
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We'll begin by proving $\sum_{n\ge0}\frac{(n+k)!}{n!k!}x^n=(1-x)^{-k-1}$ by noting the binomial theorem gives the right-hand side's $x^m$ coefficient as $\frac{(-1)^m}{m!}\prod_{j=1}^m(-k-j)$, which is exactly what we want. Applying $\sum_{k\ge0}$ gives the desired result. Note the convergence assumptions: in the first step we required $|x|<1$, while in the second we need $|1-x|>1$, in which case the result is $\frac{1}{1-x}\frac{1}{1-\frac{1}{1-x}}=-\frac{1}{x}$. For real $x$ both conditions hold iff $x\in(-1,\,0)$.
J.G.
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