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The Laguerre polynomial of degree $n$ is $$L_n(x)= \sum_{k=0}^n \frac{(-1)^k \; n\,!}{(k\,!)^2 (n-k)\;!}x^k $$

I am expanding the generating function $$\phi(x,z)= \frac{e^{-xz/(1-z)}}{1-z} $$ to get an expression involving the Laguerre Polynomials ( given $|z|<1$ ) as following $$\phi(x,z)= \sum_{k=0}^\infty L_n(x) z^n \tag{1}$$

To proceed, I just started expanding $\phi(x,z)$ but got stuck in the middle. Here is my attempt:
$$\phi(x,z)= \frac{e^{-xz/(1-z)}}{1-z}=\frac{1}{1-z}\;\;\sum_{k=0}^\infty \frac{1}{k\;!} \left(\frac{-xz}{1-z}\right)^k $$ $$=\sum_{k=0}^\infty\frac{(-1)^k}{k\;!}\frac{x^k\;z^k}{(1-z)^{1+k}} $$

Stuck from here. Probably I have to expand $(1-z)^{-(1+k)}$ but I have no idea how to do it.

How to get from here to equation $(1)$.

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1 Answer 1

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Hint

Try using the Binomial theorem $$(a+b)^k = \sum_{n=0}^k {n\choose k}a^kb^{n-k}$$ and remember the fact that $${n\choose k} = \frac{n!}{k!(n-k)!}$$

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