<s>[INST] Taylor Series for $f(z)=\frac{z}{2}+\frac{z}{e^z-1}$. The question reads as follows:

Let the function $f$ be given by $f(z)=\frac{z}{2}+\frac{z}{e^z-1}$ if
  $z\neq0$ and $f(z)=1$ if $z=0$. Show that $f$ is analytic at $z=0$ and
  that $f(z)=f(-z)$. Deduce that there is a Taylor series
  $f(z)=a_0+a_2x^2+a_4x^4+...$ valid for $|z|<2\pi$. Find $a_0,a_2,a_4$.

Clearly the function is analytic at $z=0$, and $f(z)$ is even for all $z$. Now, the singularities occur where $z=2ni\pi$ where $n\in\mathbb{Z}$. Given $|z|<2\pi$, we have that $f(z)$ is analytic within this disk. Hence a Taylor series exists for $f(z)$ such that $z_0=0$. Thus $f(z)=a_0+a_1z+a_2x^2+a_3z^3+a_4x^4+....$ in $|z|<2\pi$ where $a_n=\frac{f^{n}(0)}{n!}$ where $n\in\mathbb{Z^+}$. 
Since $f(z)$ is even this implies that only the even powers exist. Thus $f(z)=a_0+a_2x^2+a_4x^4+...$. I am having trouble finding the coefficients, $a_0,a_2,a_4$.
Thank you in advanced for you help.
 [/INST] To find the first terms of the Taylor series of $f$, there are various possibilities. One can simply differentiate the function and use $a_n = \frac{1}{n!}f^{(n)}(0)$ to obtain the coefficients, but that soon becomes unwieldy. Still, for the coefficients up to order $4$, it's doable.
Another standard way, when one has such a quotient, is to expand the numerator (trivial here) and the denominator, and divide both by a suitable power of $z$ times a constant, so that the denominator has the form $1 - h(z)$, and expand $\frac{1}{1-h(z)}$ in a geometric series. That leads to (ignoring the $\frac{z}{2}$ summand)
$$\begin{align}
\frac{z}{e^z-1} &= \frac{z}{z + \frac{z^2}{2} + \frac{z^3}{6} + \dotsb}\\
&= \frac{1}{1 + \left(\frac{z}{2} + \frac{z^2}{6} + \dotsb\right)}\\
&= 1 - \left(\frac{z}{2} + \frac{z^2}{6} + \dotsb\right) + \left(\frac{z}{2} + \frac{z^2}{6} + \dotsb\right)^2 - \dotsb\\
&= 1 - \frac{z}{2} +\left(\frac14 - \frac16\right)z^2 + \dotsb\\
&= 1 - \frac{z}{2} + \frac{z^2}{12} + \dotsb
\end{align}$$
The coefficients $a_0$ and $a_2$ are found in this way without much computation, but finding $a_4$ involves some more computation. It's still not too hard.
Yet another way is to multiply the unknown expansion with the known expansion of the denominator to obtain a recurrence for the coefficients:
$$\begin{align}
z &= \frac{z}{e^z-1}(e^z-1)\\
&= \left(\sum_{k=0}^\infty \frac{b_k}{k!}z^k\right)\cdot\left(\sum_{m=1}^\infty \frac{z^m}{m!}\right)\\
&= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1}\frac{b_k}{k!(n-k)!}\right)z^n\\
&= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1} \binom{n}{k}b_k\right)\frac{z^n}{n!}.
\end{align}$$
Equating the coefficients yields $1 = \binom{1}{0}b_0$, so $b_0 = 1$, and for $n > 1$, the recurrence
$$\sum_{k=0}^{n-1}\binom{n}{k}b_k = 0.\tag{1}$$
For $n = 2$, we obtain $0 = \binom{2}{0}b_0 + \binom{2}{1}b_1 = 1 + 2b_1$, so $b_1 = -\frac12$. For $n = 3$, it becomes
$$0 = \binom{3}{0}b_0 + \binom{3}{1}b_1 + \binom{3}{2}b_2 = 1 - \frac{3}{2} + 3b_2 \Rightarrow b_2 = \frac16.$$
Note that we have $a_k = \frac{b_k}{k!}$, so this agrees with the result above.
Using the fact that $b_{2k+1} = 0$ for $k > 0$ by the evenness of $f$, we need not compute $b_3$ and further odd coefficients, which saves some work. Computing $a_4$ with the recurrence $(1)$ is then not much work.
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