Power series expansion of a holomorphic function One must show that equality$$\sum_{n=0}^\infty n^2z^n = \frac{z(z+1)}{(1-z)^3} $$ holds for $z\in\mathbb C$ and $|z|<1$.
I tried to expand $f(z):=\frac{z(z+1)}{(1-z)^3}$ into power series where $ z_0=0$  is the center and coefficients given by $a_n=\frac{1}{2\pi i}\int_{|z|<1}\frac{f(z)}{(z-0)^{n+1}}dz$, because f is holomorphic on the open domain $D=${$z:|z|<1$}.
Is there a more easy way to show this equality?
 A: Start with $\frac{1}{1-z} = \sum_{n=0}^{+\infty} z^n$. Derive to obtain $$\frac{1}{(1-z)^2} = \sum_{n=0}^{+\infty} (n+1)z^{n} = \sum_{n=0}^{+\infty} nz^n +\frac{1}{1-z}.$$ Now derive one more time to conclude. 
A: Decompose into partial fractions:
$$
\frac{z(z+1)}{(1-z)^3}=\frac{A}{1-z}+\frac{B}{(1-z)^2}+\frac{C}{(1-z)^3}
$$
Now recall that
\begin{align}
\frac{1}{1-z}&=\sum_{n\ge0}z^n \\[4px]
\frac{1}{(1-z)^2}&=\sum_{n\ge0}(n+1)z^n \\[4px]
\frac{2}{(1-z)^3}&=\sum_{n\ge0}(n+2)(n+1)z^n
\end{align}
by differentiating.
A: We could also recall the binomial series expansion valid for $|z|<1$
\begin{align*}
\frac{1}{(1-z)^3}&=\sum_{n=0}^\infty\binom{-3}{n}(-z)^n=\sum_{n=0}^\infty\binom{n+2}{2}z^n\\
&=\frac{1}{2}\sum_{n=0}^\infty(n+2)(n+1)z^n \tag{1}
\end{align*}

Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain from (1)
\begin{align*}
[z^n]\frac{z+z^2}{(1-z)^3}&=\left([z^{n-1}]+[z^{n-2}]\right)\frac{1}{2}\sum_{k=0}^\infty(k+2)(k+1)z^k\\
&=\frac{1}{2}(n+1)n+\frac{1}{2}n(n-1)\\
&=n^2
\end{align*}

A: $\sum_{n \ge 0} n^2z^n$ is absolutely convergent for $|z| < 1$ and so the following manipulations are valid for $|z| < 1$ :
$(1-z)^3(z+4z^2+9z^3+16z^4+\ldots) 
\\ = (1-z)^2(z-z^2+4z^2-4z^3+9z^3-9z^4+16z^4+\ldots) 
\\ = (1-z)^2(z+3z^2+5z^3+7z^4+\ldots)
\\ = (1-z)(z-z^2+3z^2-3z^3+5z^3-5z^4+7z^4+\ldots)
\\ = (1-z)(z+2z^2+2z^3+2z^4+\ldots)
\\ = z-z^2+2z^2-2z^3+2z^3-2z^4+2z^4+\ldots
\\ = z+z^2
$
This proves that $z+4z^2+9z^3+16z^4+\ldots = \frac {z+z^2}{(1-z)^3}$ when $|z| <1 $ (and $1-z \neq 0$ but this check is redundant)
