calculate the series $\sum_{n=0}^\infty n(n+1)z^n $

Question

I have found the convergence domain, which is $$|z|<1$$ now for this domain I am trying to calculate the series $$\sum_{n=0}^\infty n(n+1)z^n $$

Answer 1

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*}

Answer 2

A calculus-free alternative: note the $n=0$ term vanishes so, by the binomial theorem, the dummy variable $m:=n-1$ gives$$\begin{align}\sum_{n\ge0}n(n+1)z^n&=2z\sum_{m\ge0}\binom{m+2}{m}z^m\\&=2z\sum_m\binom{-3}{m}(-z)^m\\&=2z(1-z)^{-3}.\end{align}$$

Answer 3

I thought it might be instructive to present an approach that relies on elementary pre-calculus tools only. To that end, we proceed.

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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!