Get closed form of a series from its generating function

Question

If we have the ordinary generating function of a series:

$f(x) = \frac{1}{e^z -3}$

Can we get its closed form?

Answer 1

I think we can find the terms of the sequence using the geometric series and the power series expansion of $e^x$. \begin{align*} \frac{1}{e^x - 3} &= -\frac{1}{3} \frac{1}{1 - e^x/3} = -\frac{1}{3} \sum_{n \geq 0} \left(\frac{e^x}{3}\right)^n = -\frac{1}{3} \sum_{n \geq 0} \frac{1}{3^{n}} e^{nx} = -\sum_{n \geq 0} \frac{1}{3^{n+1}} \sum_{k \geq 0} \frac{(nx)^k}{k!}\\ &= -\frac{1}{3} \sum_{k \geq 0} \left(\sum_{n \geq 0} \frac{n^k}{3^{n}}\right) \frac{1}{k!} x^k \, . \end{align*} We simplify the series $\sum_{n \geq 0} n^k x^n$ by repeatedly differentiating the series $\sum_{n \geq 0} x^n = \frac{1}{1-x}$ and multiplying by $x$. That is, $$ \sum_{n \geq 0} n^k x^n = \left(x \frac{d}{dx}\right)^k \frac{1}{1-x} \, . $$ (There may be a simpler expression for this, but I'm not aware of one.) By setting $x = 1/3$, we obtain the sum in parentheses above. Thus the $k^\text{th}$ term of your sequence will be the coefficient of $x^k$, which is \begin{align*} -\frac{1}{3} \sum_{k \geq 0} \left(\sum_{n \geq 0} \frac{n^k}{3^{n}}\right) \frac{1}{k!} &= -\frac{1}{3} \sum_{k \geq 0} \left[\left(x \frac{d}{dx}\right)^k \frac{1}{1-x}\right]_{x = 1/3} \frac{1}{k!} \, . \end{align*} I guess this is a closed-ish form.

Answer 2

The coefficients of many generating functions just don't have "simple" form, when they turn out often they get christened as special sequences:

• Bell numbers (total number of partitions of $n$ elements): $e^{e^z - 1}$
• Ordered Bell numbers (number of ordered partitions of $n$ elements, "competitions with ties"): $\frac{1}{2 - e^z}$
• Bernoulli numbers: $\frac{z e^z}{e^z - 1}$

This is just a list off the top of my head. You might get an expression for your sequence in terms of those "well known" sequences.

In any case, OEIS should be your first stop.