cepheid said:
Does anyone have tips on how to sum the following series?
\sum_{n=1}^{\infty} n^2 w^n
Region of convergence is for |w| < 1
(1) \ \ \ \ z \ = \ \sum_{n=0}^{\infty} w^{n} \ = \ (1 \ - \ w)^{-1} \ =
(2) \ \ \ \ \ \ \ \ \ \ = \ 1 \ + \ w \ + \ \sum_{n=2}^{\infty} w^{n}
3 \ \ \ \ \ \ \frac {dz} {dw} \ = \ \left ( 1 \ - \ w \right )^{-2} =
(4) \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ 1 \ \ + \ \ \sum_{n=2}^{\infty} n \cdot w^{n-1} \ = \ 1 \ \ + \ \ w^{-1} \cdot \sum_{n=2}^{\infty} n \cdot w^{n}
(5) \ \ \ \ \ \Longrightarrow \ \sum_{n=2}^{\infty} n \cdot w^{n} \ = \ w \cdot \left( \left(1 \ - \ w \right)^{-2} \ - \ 1 \ \right)
6 \ \ \ \ \ \ \frac {d^{2}z} {dw^{2}} \ = \ 2 \cdot \left ( 1 \ - \ w \right )^{-3} \ =
(7) \ \ \ \ \ \ \ \ \ \ = \ \sum_{n=2}^{\infty} n \cdot ( n \ - \ 1 ) \cdot w^{n-2} \ =
(8) \ \ \ \ \ \ \ \ \ \ = \ w^{-2} \cdot \sum_{n=2}^{\infty} n \cdot ( n \ - \ 1 ) \cdot w^{n} \ =
(9) \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ w^{-2} \cdot \left ( \sum_{n=2}^{\infty} n^{2} \cdot w^{n} \ - \ \sum_{n=2}^{\infty} n \cdot w^{n} \right ) \ =
(10) \ \ \ \ \ \ \ \ \ \ \ = \ w^{-2} \cdot \left ( \sum_{n=2}^{\infty} n^{2} \cdot w^{n} \ - \ w \cdot \left( \left(1 \ - \ w \right)^{-2} \ - \ 1 \ \right) \right ) \
(11) \ \ \ \ \color{red} \Longrightarrow \ \sum_{n=2}^{\infty} n^{2} \cdot w^{n} \ = \ 2 \cdot w^{2} \cdot \left ( 1 \ - \ w \right )^{-3} \ \ + \ \ w \cdot \left( \left(1 \ - \ w \right)^{-2} \ - \ 1 \ \right)
(12) \ \ \ \ \color{red} \Longrightarrow \ \sum_{n=1}^{\infty} n^{2} \cdot w^{n} \ \ = \ \ w \ \ + \ \ 2 \cdot w^{2} \cdot \left ( 1 \ - \ w \right )^{-3} \ \ + \ \ w \cdot \left( \left(1 \ - \ w \right)^{-2} \ - \ 1 \ \right)
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