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Using the identity $\frac{1}{1-z} = 1 + z + z^2 + \ldots$ for $|z| < 1$, find closed forms for the sums $\sum n z^n$ and $\sum n^2 z^n$.

My solution: Because $\displaystyle1 + z + z^{2} + \ldots = \frac{1}{1-z}$, $\displaystyle1 + 2z + 3z^2 + \ldots = \sum_{n=1}^\infty n z^{n-1} = \frac{1}{(1-z)^2}$ and $\displaystyle\sum_{n=1}^\infty n z^n = \frac{z}{(1-z)^2}.$ Similarly, $\displaystyle\sum_{n=1}^\infty n^2 z^{n-1}=\frac{1+z}{(1-z)^3}$ and $\displaystyle\sum_{n=1}^\infty n^2 z^n=\frac{z(1+z)}{(1-z)^3}$.

Could you please improve this exercise?

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    I would make explicit the differentiation step in your first instance (and show the algebra on the derivative in the second) - right now you simply say '$\Sigma_n z^n=$ this, so $\Sigma_n nz^{n-1}=$ this' without explaining how you get from a) to b). Other than that, it looks fairly clear!2012-05-01

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I'd do the second one like this $ \begin{align} \sum_{n=1}^\infty n^2 z^{n-1} & = z\sum_{n=2}^\infty n(n-1) z^{n-2} + \sum_{n=1}^\infty nz^{n-1} \\[10pt] & = z\frac{d^2}{dz^2} \frac{1}{1-z} + \frac{d}{dz} \frac{1}{1-z}= \cdots\cdots \end{align} $ (and then multiply by $z$).

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    And you can use this method to simplify all power series with coefficient a polynomial of the degree~~2012-05-02