Calculating Convergent Series: Tips for $\sum_{n=1}^{\infty} n^2 w^n$

AI Thread Summary
To sum the series ∑_{n=1}^{∞} n^2 w^n, the region of convergence is defined for |w| < 1. A useful approach is to differentiate the geometric series, starting with z = ∑_{n=0}^{∞} w^n = (1 - w)^{-1}. By differentiating this series, one can derive expressions for ∑_{n=2}^{∞} n w^n and subsequently for ∑_{n=2}^{∞} n^2 w^n. The final result for the series is given by ∑_{n=1}^{∞} n^2 w^n = w + 2w^2(1 - w)^{-3} + w((1 - w)^{-2} - 1).
cepheid
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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
 
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Try to exploit it's similarity with a geometric series.
Hint: Differentiate.
 
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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Just Wanted To Know How To Make The Summation And Powers
 
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