The exercise
Compute the Maclaurin expansion of $\frac{1 - x \cosh(a)}{1 - 2 \cosh(a) x + x^2}$ at $0$ for any order $n \in \mathbb{N}$.
My try
I remarked $1 - 2 \cosh(a) x + x^2 = (x-e^a)(x-e^{-a})$, not sure it helps.
One also has, setting $u(x) := x^2 - 2 x \cosh(a)$, $\frac{1}{1-u(x)} = 1 + u(x) + u(x)^2 + \dots + u(x)^n + o(u(x))$ as $u \underset{0}{\rightarrow} 0$. We also have $o(u(x)) = o(x)$ when $x \to 0$.
I could compute the expansion for $n = 1, 2, 3...$ but having a clean formula for any $n$ seems hard to me.
I am almost sure there is a trick that would ease a lot the resolution of the exercise.
Thanks for the help.
Your fraction is$$\frac{1-x(e^a+e^{-a})/2}{(x-e^a)(x-e^{-a})}=\frac12\sum_\pm\frac{1}{1-e^{\pm a}x}=\sum_{n\ge0}x^n\cosh na.$$