# What is the $n$-th sequence element for the generating function $\frac{1}{(1-ax)^2}$?

for e.g. for $\frac{1}{(1-ax)} = a^n$ or for $\frac{1}{(1-x)^2} = n+1$

generating function = $\frac{1}{(1-ax)^2}$

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You are looking at the derivative of $1/(1-ax)$ with respect to $x$, divided by $a$. Hence the $x^n$ term in this series is... –  Did Nov 10 '11 at 8:02
(-2)/(1-ax)^2 ? –  user494461 Nov 10 '11 at 8:09
Hmmm... The $x^n$ term in a series ought to be $a_nx^n$ for a given coefficient $a_n$, depending (possibly) on $n$ but not on $x$. Anyway, in the meantime you received full answers, hence, unless you wish to ask precise questions on specific steps of these, let us stop here. –  Did Nov 10 '11 at 10:08

You give that $\frac{1}{(1-x)^2}$ is the generating function of $a_n=n+1$. That means that $$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k\tag{1}$$ Therefore, $$\frac{1}{(1-ax)^2}=\sum_{k=0}^\infty (k+1)(ax)^k\tag{2}$$ Thus, $\frac{1}{(1-ax)^2}$ is the generating function for $a_n=(k+1)a^k.$

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Method 1:

Use the general formula for the $k$-th derivative of a power (or establish through induction):

$$\frac{\mathrm d^k}{\mathrm dx^k}\frac1{(1-ax)^2}=\frac{(k+1)!a^k}{(1-ax)^{k+2}}$$

Method 2:

Since $\dfrac1{(1-ax)^2}=\dfrac1{(1-ax)}\cdot\dfrac1{(1-ax)}$, and $\dfrac1{(1-ax)}$ is the generating function for $a^k$, the coefficients of $\dfrac1{(1-ax)^2}$ are given by the autoconvolution of $a^k$; i.e. the sum

$$\sum_{j=0}^k a^j a^{k-j}$$

whose simplification I leave to you.

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Another method. By the binomial theorem: $$(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k$$ Now we have, using Knuth's notation for falling/rising factorials: $m^{\underline{k}} = m (m - 1) \ldots (m - k + 1)$ and $m^{\overline{k}} = m (m + 1) \ldots (m + k - 1)$: $$\binom{-m}{k} = \frac{(-m)^{\underline{k}}}{k!} = \frac{(-1)^k m^{\overline{k}}}{k!} = \frac{(-1)^k (m + k - 1)^{\underline{k}}}{k!} = (-1)^k \binom{m + k - 1}{k} = (-1)^k \binom{k + m - 1}{m - 1}$$ Note that this is just a $m - 1$-th degree polynomial in $k$.

So we have: $$(1 - u)^{-m} = \sum_{k \ge 0} \binom{k + m - 1}{m - 1} u^k$$ The rest of the fun I leave to you.

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