Find the power series representation of the function

Question

I have to find $a_n$ so, my first attempt was to do partial fraction, but with no real solution. Any recommendation will be appreciate $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$

Answer 1

Let us replace $x$ with $z+1$ for the sake of simplicity.
We have to find the sequence $\{a_n\}_{n\geq 0}$ given by: $$ f(z)=\frac{1}{(2z-1)(z^2+z+1)}=\sum_{n\geq 0}a_n z^n \tag{1} $$ but $f(z)$ is a meromorphic function with simple poles at $z=\frac{1}{2},\,z=\omega,\,z=\omega^2$, where $\omega=e^{\frac{2\pi i}{3}}$.
By computing the residues of $f(z)$ at such points we get a partial fraction decomposition for $f(z)$:

$$ f(z) = \frac{2}{7}\cdot \frac{1}{x-\frac{1}{2}}-\frac{1}{7}\cdot\frac{3-x-2x^2}{1-x^3} \tag{2}$$ and by performing expansions as geometric series it follows that: $$ a_n = -\frac{4}{7} 2^n-\frac{1}{7}\cdot\left\{\begin{array}{rcl}3 & \text{if} & n\equiv 0\pmod{3}\\ -1 & \text{if} & n\equiv 1\pmod{3}\\ -2 & \text{if} & n\equiv 2\pmod{3}\end{array}\right.\tag{3}$$ that can be stated as:

$$ a_n \text{ is the closest integer to } -\frac{2^{n+2}}{7}. \tag{4}$$

Answer 2

Note that the geometric series is valid also in the complex field $$ \frac{1} {{1 - z}} = \sum\limits_{0\, \leqslant \,n} {z^{\,n} } \quad \left| \begin{gathered} \;z \in \;\mathbb{C}\; \hfill \\ \,\left| z \right| < 1 \hfill \\ \end{gathered} \right. $$

and that the partial fraction decomposition can be operated also in the complex field, so that in your case we can write

$$ \begin{gathered} f(z,\omega ) = \frac{1} {{\left( {2z - 3} \right)\left( {z - \omega } \right)\left( {z + \omega } \right)}} = \hfill \\ = \frac{1} {{2\left( {z - 3/2} \right)\left( {z - \omega } \right)\left( {z + \omega } \right)}} = \hfill \\ = - \frac{1} {{3\left( {3/2 - \omega } \right)\left( {3/2 + \omega } \right)\left( {1 - 2/3\,z} \right)}} + \frac{1} {{4\omega ^2 \left( {3/2 - \omega } \right)\left( {1 - z/\omega } \right)}} + \frac{1} {{4\omega ^2 \left( {3/2 + \omega } \right)\left( {1 - \left( { - z/\omega } \right)} \right)}} \hfill \\ \end{gathered} $$

and the expression for $a_n$ follows easily.