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The power series is: $$\sum_{n=0}^\infty n^2 z^n$$

to which I have to find the function $f(z)$.

Taking the derivative or integrating did not get me anywhere in making it look like a geometric series.

Any help would be awesome.

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    Consider the map $g(z) \mapsto z\cdot g'(z)$. What does that do to a convergent power series about $0$? – 2017-02-18
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    sorry I don't follow you – 2017-02-18
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    If $$g(z) = \sum_{n = 0}^{\infty} a_n z^n,$$ then what is $z\cdot g'(z)$? – 2017-02-18
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    it's equal to $ng(z)$ – 2017-02-18
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    That's if $a_n = n^2$. I'm looking at a generic power series. The goal is to reach your $f$ from a known function through that operation. – 2017-02-18
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    then it is $\sum_{n=1}^\infty n a_n z^n$ – 2017-02-18
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    Right. Now, from what power series do we reach $\sum n^2 z^n$ with this operation? – 2017-02-18
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    so that power series is $\sum n z^n$. if we let $g = \sum n z^n$, then $zg'(z) = \sum n^2 z^n$ and the function corresponding to g(I have found before) is $z/(1-z)^2$. so whats left is to take the derivative of this function nd multiply by z – 2017-02-18
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    is that okay..? – 2017-02-18
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    Yes. Differentiating may be more convenient using $\dfrac{z}{(1-z)^2} = \dfrac{1 - (1-z)}{(1-z)^2} = \dfrac{1}{(1-z)^2} - \dfrac{1}{1-z}$. – 2017-02-18
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    got it, thank you very much – 2017-02-18

2 Answers 2

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Since $$ \dfrac{1}{1-z}=\sum_{n=0}^\infty z^n \quad \forall |z|<1, $$ we have \begin{eqnarray} \sum_{n=1}^\infty n^2z^n&=&z\sum_{n=1}^\infty n^2z^{n-1}=z\dfrac{d}{dz}\left(\sum_{n=1}^\infty nz^n\right)=z\dfrac{d}{dz}\left(z\sum_{n=1}^\infty nz^{n-1}\right)=z\dfrac{d}{dz}\left[z\dfrac{d}{dz}\left(\sum_{n=0}^\infty z^n\right)\right]\\ &=&z\dfrac{d}{dz}\left[z\dfrac{d}{dz}\left(\dfrac{1}{1-z}\right)\right]=z\dfrac{d}{dz}\left[\dfrac{z}{(z-1)^2}\right]=z\dfrac{(z-1)^2-2z(z-1)}{(z-1)^4}\\ &=&z\dfrac{z-1-2z}{(z-1)^3}=\dfrac{z(1+z)}{(1-z)^3} \end{eqnarray}

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\begin{eqnarray*} \sum_{n=0}^{\infty} z^n =\frac{1}{1-z} \end{eqnarray*} Differentiate this & we get \begin{eqnarray*} \sum_{n=0}^{\infty}n z^{n-1} =\frac{1}{(1-z)^2} \end{eqnarray*} Diffentiate again \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{n(n-1)}{2}z^{n-2} =\frac{1}{(1-z)^3} \end{eqnarray*} Now shift the sum index by one \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{n(n+1)}{2}z^{n-2} =\frac{z}{(1-z)^3}. \end{eqnarray*} Multiply the second to last equation by $z^2$ \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{n(n+1)}{2}z^{n} =\frac{z^2}{(1-z)^3}. \end{eqnarray*}

Now add the previous two equations and shift the index by 1 \begin{eqnarray*} \sum_{n=0}^{\infty} n^2 z^{n} =\frac{z(1+z)}{(1-z)^3} \end{eqnarray*}