calculate the series $\sum_{n=0}^\infty n(n+1)z^n $ I have found the convergence domain, which is $$|z|<1$$
now for this domain I am trying to calculate the series $$\sum_{n=0}^\infty n(n+1)z^n $$
 A: A calculus-free alternative: note the $n=0$ term vanishes so, by the binomial theorem, the dummy variable $m:=n-1$ gives$$\begin{align}\sum_{n\ge0}n(n+1)z^n&=2z\sum_{m\ge0}\binom{m+2}{m}z^m\\&=2z\sum_m\binom{-3}{m}(-z)^m\\&=2z(1-z)^{-3}.\end{align}$$
A: 
I thought it might be instructive to present an approach that relies on elementary pre-calculus tools only.  To that end, we proceed.


First, we note that $n(n+1)=2\sum_{k=0}^n k$.  Hence, we have
$$\begin{align}
\sum_{n=0}^\infty n(n+1)z^n&=2\sum_{n=0}^\infty \sum_{k=0}^n kz^n\\\\
&=2\sum_{k=0}^\infty k\sum_{n=k}^\infty z^n\\\\
&=2\sum_{k=0}^\infty k \,\frac{z^k}{1-z}\\\\
&=2\sum_{k=1}^\infty k \,\frac{z^k}{1-z}
\end{align}$$
Next, note that $k=\sum_{n=1}^{k}(1)$.  Hence, we have
$$\begin{align}
2\sum_{k=1}^\infty k \,\frac{z^k}{1-z}&=\frac2{1-z}\,\sum_{k=1}^\infty \sum_{n=1}^{k}z^k\\\\
&=\frac2{1-z}\,\sum_{n=1}^{\infty}\sum_{k=n}^\infty z^k\\\\
&=\frac2{1-z}\,\sum_{n=1}^{\infty} \frac{z^n}{1-z}\\\\
&=\frac{2z}{(1-z)^3}
\end{align}$$
And we are done!
A: Start with
\begin{eqnarray*}
\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}.
\end{eqnarray*}
Differentiate this twice & mutliply by $z$.
\begin{eqnarray*}
\sum_{n=0}^{\infty} n(n+1) z^n = \frac{2z}{(1-z)^3}.
\end{eqnarray*}
