Power series expansion of $\frac{z}{(z^3+1)^2}$ around $z=1$ I want to expand $f(z)=\frac{z}{(z^3+1)^2}$ around $z=1$. That is, I want to find the coefficients $c_n$ such that $f(z) = \sum_{n=0}^\infty c_n (z-1)^n$.
So far, my first strategy was using long division after I expanded both the numerator and the denominator about $z=1$. To this end, we first apply a substitution $u:= z-1$, after we'll expand around $u=0$.
For the denominator we simply get $$1 + u$$.
For the numerator we have 
$$
\frac{1}{\big((u+1)^3+1\big)^2} = \frac{1}{2^2} \frac{1}{\big(1 + \frac{u^3+3u(u+1)}{2}\big)^2}  \quad \mbox{put $x:= \frac{u^3+3u(u+1)}{2}$}\\
=-\frac{1}{2^2}\frac{\partial d}{\partial x}\frac{1}{x+1} \\
= -\frac{1}{2^2}(-1+2x-3x^2 +\ \ldots\ ) \\
=-\frac{1}{2^2}\big(1+6u-21u^2-52u^3+\ \ldots \ \big)
$$
After using long division, we could try to see a pattern in the coefficients of $u^n$ but this doesn't seem to bear any quick results.

My second attempt was using the binomial theorem to expand the power series of $$f(z) = \sum_{n=0}^\infty a_n z^{3n+1} \quad \mbox{with} \quad a_n = (n+1)(-1)^n$$ about $z=1$.
To this end, we need to reorganize the $u^n$ in the following sum
$$
\sum_{n=0}^\infty (u+1)^n = \sum_{n=0}^\infty \sum_{k=0}^n \binom nk u^k .
$$
This seems fairly laborious. Below we have the terms for increasing $n$.
$$
1 \\
1 + u \\
1 + 2u + u^2 \\
1 + 3u + 3u^2 + u^3 \\
 \vdots \\ 
1 + mu + \binom m2 + \ldots + mu^{m-1} + u^m \\
1 + (m+1)u + \binom{m+1}{2}u^2 + \ldots + (m+1)u^m + u^{m+1} \\
1 + (m+2)u + \binom{m+2}{2}u^2 + \ldots + (m+2)u^{m+1} + u^{m+2} \\
\vdots
$$
If we look at the terms of $1$, $u$ and $u^2$, their coefficients $b_n$ in the sum over $n$ suggest the following coefficients for $1$, $u$ and $u^2$ respectively.
$$
b_0 = \sum_{n=0}^\infty 1  \\
b_1 =\sum_{n=0}^\infty n  \\
b_2 = \sum_{n=0}^\infty \binom n2 \ .   
$$  
Expanding on this, we arrive at
$$
\sum_{i=0}^\infty b_i u^i = \sum_{i=0}^\infty \bigg( \sum_{k=0}^\infty \binom  ki \bigg)u^i
$$
, which seems very incorrect, since we have a double infinite sum.
 A: Notice that
$$\frac z{(1+z^3)^2}=-z\frac{\partial}{\partial z^3}\frac1{1+z^3}=-z\frac\partial{\partial z^3}\sum_{n=0}^\infty(-1)^n(z^3)^n=\sum_{n=0}^\infty n(-1)^{n+1}z^{3n-2}$$
This expansion works for $|z|<1$, and since we know the expansion exists at $z=1$, we may apply the method described in this answer to get the expansion at $z=1$.
$$=\sum_{j=0}^\infty \left(\sum_{k=j}^\infty \binom kj a_k \right)(z-1)^j$$
where $a_{3n+1}=(-1)^n(n+1)$ for $n\in\mathbb N$, else $a_k=0$.
However, note that this method is usually only viable when the center for the radius of convergence was within the radius of convergence of the original series.  For example, plugging $z=1$ into our "expansion" yields
$$\frac14\stackrel?=1-2+3-4+5-6+\dots$$
However, an interesting case happens that we may regularize our divergent series to equal our original function by applying an Euler sum to it:
$$f(z)=\sum_{j=0}^\infty\sum_{k=\lfloor j/3\rfloor}^\infty \binom{3k+1}j (-1)^k(k+1)(z-1)^j\\=\sum_{j=0}^\infty\sum_{i=\lfloor j/3\rfloor}^\infty\frac1{2^i}\sum_{k=\lfloor j/3\rfloor}^i\binom ik\binom{3k+1}j (-1)^k(k+1)(z-1)^j$$
which now converges as desired.
