Generating Function explicit formula

Question

Say i got $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$

I used $\displaystyle{\frac{1}{(1+3x)}}$ $=\sum_{n=0}^\infty(-3)^n x^n$ and differentiated twice

I got $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ = $=\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$

Multiply (1-2x) on both side I got

= $\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$ - $2\sum_{n=0}^\infty [((n-1)(n-2)(-3)^{(n-1)}))/18] x^{(n-2)}$

$=\sum_{n=0}^\infty [(5n-4)(n-1)(-3)^n /54 ] x^{(n-2)}$

Is that correct ? I had a feeling that its wrong...

Coefficient of $z^{(n-2)}$ is $\displaystyle{\frac{(5n-4)(n-1)(-3)^n}{54}}$?

Answer 1

You have a small mistake in the second line: $\frac{1-2x}{1+3x}=(1-2x)\overset{\infty}{\underset{n=0}{\sum}}(-3)^nx^n$.
But you can use the following: for all $a \in \mathbb{C}$ we have $(1+x)^a=\overset{\infty}{\underset{n=0}{\sum}}\binom{a}{n}x^n$, where $\binom{a}{n}$ is the formal notation for $\frac{a(a-1)\cdot...\cdot(a-n+1)}{n!}$ and $\binom{a}{0}=1$. Applying this, you can check your answer.

Answer 2

Adding on Dennis Gulko's answer.

For any $m \in \mathbb{N}$ fixed, it holds $$ \frac{1}{{(1 + mx)^m }} = \sum\limits_{n = 0}^\infty {\frac{{m \cdots (m + n - 1)}}{{n!}}( - m)^n x^n }, $$ provided that $|x|$ is sufficiently small.

This fact (confirmed numerically) can be shown in a probabilistic setting as follows. Let $X$ be a gamma$(m,1)$ random variable, so that $X$ has density function $f(x)=e^{ - x} x^{m - 1} /\Gamma (m)$, $x > 0$. The moment-generating function (MGF) of $X$ is given by $$ {\rm E}[e^{tX} ] = \frac{1}{{(1 - t)^m }} $$ (indeed, note that $1/(1-t)$ is the MGF of the exponential$(1)$ distribution, and $X$ can be written as a sum of $m$ independent exponential$(1)$ variables). Further, the $n$th moment ($n=0,1,2,\ldots$) of $X$ is given by $$ \mu_n' = \int_0^\infty {x^n f(x)\,{\rm d}x} = \frac{1}{{\Gamma (m)}}\int_0^\infty {x^{n + m - 1} e^{ - x} \,{\rm d}x} = \frac{{\Gamma (n + m)}}{{\Gamma (m)}} = m \cdots (m + n - 1) $$ (note that $\mu_0' = 1$). Hence, for all $t$ in a neighborhood of $0$, it holds $$ {\rm E}[e^{tX} ] = \sum\limits_{n = 0}^\infty {\frac{{\mu _n' }}{{n!}}t^n } = \sum\limits_{n = 0}^\infty {\frac{{m \cdots (m + n - 1)}}{{n!}}t^n } . $$ Finally, putting $t = -mx$, we get $$ \frac{1}{{(1 + mx)^m }} = \sum\limits_{n = 0}^\infty {\frac{{m \cdots (m + n - 1)}}{{n!}}( - m)^n x^n } . $$