Taylor series for $f(z)=z\cos(z)/(z^2-2)$ Let $f(z)=z\cos(z)/(z^2-2)$, $z\in\mathbb{C}$.
Show that the Taylor series for $f$ around $0$ is given by
$$f(z)=\sum_{k=0}^\infty a_{2k+1} z^{2k+1} $$
with $a_1=-1/2$ and $a_{2k+1}=(a_{2k-1}+(-1)^{k+1}/(2k)!)/2$ for $k\geq 1$.

I tried to use the power series for $\cos(z)$ and the Taylor series for $z/(z^2-2)$ and somehow combine them, but got stuck.
This seems like it should be simple enough, but I can't get anywhere. Any help is appreciated


We observe since $\cos(z)=\sum_{j=0}^\infty \frac{(-1)^j}{(2j)!}z^{2j}$ is an even function, the function $f(z)$
  is odd which makes the series representation
  \begin{align*}
f(z)=\frac{z\cos(z)}{z^2-2}=\sum_{k=0}^\infty a_{2k+1}z^{2k+1}\tag{1}
\end{align*}
  plausible.

It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance
\begin{align*}
[z^{2k+1}]f(z)&=[z^{2k+1}]\sum_{j=0}^\infty a_{2j+1}z^{2j+1}=a_{2k+1}
\end{align*}
We multipy (1) with $z^2-2$ and consider
\begin{align*}
f(z)(z^2-2)=z\cos(z)
\end{align*}

We obtain from (1) for $k\geq 0$:
  \begin{align*}
[z^{2k+1}]f(z)(z^2-2)&=[z^{2k-1}]f(z)-2[z^{2k+1}]f(z)\\
&=a_{2k-1}-2a_{2k+1}
\end{align*}

Here we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

On the other hand we obtain from (1) for $k\geq 0$:
  \begin{align*}
[z^{2k+1}]z\cos(z)=[z^{2k}]\cos(z)=\frac{(-1)^k}{(2k)!}
\end{align*}
Equating both results gives
  \begin{align*}
\color{blue}{a_{2k+1}=\frac{1}{2}\left(a_{2k-1}-\frac{(-1)^{k}}{(2k)!}\right)}
\end{align*}
  and the claim follows.
We also obtain from (1)
  \begin{align*}
[z^1]f(z)(z^2-2)&=-2[z^1]f(z)\\
&=\color{blue}{-2a_1}\\
[z^1]z\cos(z)&=[z^0]\cos(z)\\
&\color{blue}{=1}
\end{align*}
  from which $\color{blue}{a_1=-\frac{1}{2}}$ follows.

