Is there an easier way of finding a Taylor series than just straight computing the formula? Let's say the task is to find the Taylor series at the origin of the function $$f(x) = \frac{3x}{1-x-2x^2}$$
The formula is $T^n_0 =\sum^n_{k=0} \frac{f^{(k)}(0)}{k!}x^k$. If I follow this formula, I need to at least compute the 4th or 5th derivatives to find the general expression which takes a long time and very likely ends up with a wrong expression.
 A: There’s another way, not as good as applying what you know about geometric series and others of your acquaintance. But it’s completely effective.
Have you learned how to divide polynomials? It’s taught in high school here in the States. To get a series expansion for a rational function, do a long division in the same spirit, but simply arrange your polynomials in ascending order of degree rather than descending, as you did in high school. Just watch out for careless errors, since they can badly spoil your result.
A: HINT: Note that $1-x-2x^2=(1-2x)(1+x)$, so
$$\frac{3x}{1-x-2x^2}=\frac1{1-2x}-\frac1{1+x}\;.$$
Now use a known Taylor series expansion to expand each of the fractions on the right-hand side, and combine the series into a single series.
Added: Let me emphasize Math$1000$’s comment below: once you know a few power series, you should always try to make use of them to get new ones. Here it’s just a matter of adding a couple, but sometimes you may have to work a bit harder: differentiating or integrating one, multiplying it by some power of $x$, or applying some combination of these manipulations.
A: Solution
$\dfrac{3z}{1-z-2z^2} = 3z \sum_{k=0}^\infty (z+2z^2)^k$ as $z \to 0$ since $z+2z^2 \to 0$. [So there is power series.]
Let $(a_k:k\in\mathbb{N})$ be a sequence such that $\dfrac{3z}{1-z-2z^2} = \sum_{k=0}^\infty a_k z^k$.
Then $3z = (1-z-2z^2) \sum_{k=0}^\infty a_k z^k = a_0 + (a_1-a_0) z + \sum_{k=0}^\infty (a_{k+2}-a_{k+1}-2a_k) z^{k+2}$
Thus $a_0 = 0$ and $(a_1-a_0) = 3$ and $a_{k+2}-a_{k+1}-2a_k = 0$ for all $k \in \mathbb{N}_{\ge 2}$.
Notes
This method works even when the denominator is a large polynomial and partial fraction decomposition would be ridiculous to do by hand. However, if you want a closed-form solution you still need to solve the recurrence, which would give exactly the same answer. If you look carefully this also shows the relationship between partial fraction decomposition (including the case with multiple roots) and recurrence relations.
