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One must show that equality$$\sum_{n=0}^\infty n^2z^n = \frac{z(z+1)}{(1-z)^3} $$ holds for $z\in\mathbb C$ and $|z|<1$.

I tried to expand $f(z):=\frac{z(z+1)}{(1-z)^3}$ into power series where $ z_0=0$ is the center and coefficients given by $a_n=\frac{1}{2\pi i}\int_{|z|<1}\frac{f(z)}{(z-0)^{n+1}}dz$, because f is holomorphic on the open domain $D=${$z:|z|<1$}.

Is there a more easy way to show this equality?

  • 0
    Examine the expansion $$ {1 \over 1 - z} = 1 + z + z^2 + \ldots, $$ and take its cube using the Cauchy product https://en.wikipedia.org/wiki/Cauchy_product twice. – 2017-02-07
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    Start with $\sum z^n = 1/(1-z)$ and differentiate a few times – 2017-02-07
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    In general, we have the [polylogarithm](https://en.wikipedia.org/wiki/Polylogarithm) $$\text{Li}_s(z)=\sum_{n=1}^\infty\frac{z^n}{n^s}$$ – 2017-02-07

4 Answers 4

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Start with $\frac{1}{1-z} = \sum_{n=0}^{+\infty} z^n$. Derive to obtain $$\frac{1}{(1-z)^2} = \sum_{n=0}^{+\infty} (n+1)z^{n} = \sum_{n=0}^{+\infty} nz^n +\frac{1}{1-z}.$$ Now derive one more time to conclude.

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Decompose into partial fractions: $$ \frac{z(z+1)}{(1-z)^3}=\frac{A}{1-z}+\frac{B}{(1-z)^2}+\frac{C}{(1-z)^3} $$

Now recall that \begin{align} \frac{1}{1-z}&=\sum_{n\ge0}z^n \\[4px] \frac{1}{(1-z)^2}&=\sum_{n\ge0}(n+1)z^n \\[4px] \frac{2}{(1-z)^3}&=\sum_{n\ge0}(n+2)(n+1)z^n \end{align} by differentiating.

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We could also recall the binomial series expansion valid for $|z|<1$ \begin{align*} \frac{1}{(1-z)^3}&=\sum_{n=0}^\infty\binom{-3}{n}(-z)^n=\sum_{n=0}^\infty\binom{n+2}{2}z^n\\ &=\frac{1}{2}\sum_{n=0}^\infty(n+2)(n+1)z^n \tag{1} \end{align*}

Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain from (1)

\begin{align*} [z^n]\frac{z+z^2}{(1-z)^3}&=\left([z^{n-1}]+[z^{n-2}]\right)\frac{1}{2}\sum_{k=0}^\infty(k+2)(k+1)z^k\\ &=\frac{1}{2}(n+1)n+\frac{1}{2}n(n-1)\\ &=n^2 \end{align*}

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$\sum_{n \ge 0} n^2z^n$ is absolutely convergent for $|z| < 1$ and so the following manipulations are valid for $|z| < 1$ :

$(1-z)^3(z+4z^2+9z^3+16z^4+\ldots) \\ = (1-z)^2(z-z^2+4z^2-4z^3+9z^3-9z^4+16z^4+\ldots) \\ = (1-z)^2(z+3z^2+5z^3+7z^4+\ldots) \\ = (1-z)(z-z^2+3z^2-3z^3+5z^3-5z^4+7z^4+\ldots) \\ = (1-z)(z+2z^2+2z^3+2z^4+\ldots) \\ = z-z^2+2z^2-2z^3+2z^3-2z^4+2z^4+\ldots \\ = z+z^2 $

This proves that $z+4z^2+9z^3+16z^4+\ldots = \frac {z+z^2}{(1-z)^3}$ when $|z| <1 $ (and $1-z \neq 0$ but this check is redundant)