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Given a series $\sum_{i > 0}\frac{i^2}{z^i}$, and $\sum_{i > 0}\frac{i}{z^i} = \frac{z}{(z - 1)^2}$ I need to find the sum

My method does not require differentiation but there is a difficulty.

Let $S = \frac{1^2}{z} + \frac{2^2}{z^2} + \frac{3^2}{z^3} + ... + \frac{i^2}{z^i}$

Let $zS = 1 + \frac{2^2}{z} + \frac{3^2}{z^2} + ... + \frac{i^2}{z^{i - 1}}$

Thus, $zS - S = 1 + \frac{2^2 - 1^2}{z} + \frac{3^2 - 2^2}{z^2} + ... + \frac{i^2 - (i - 1)^2}{z^{i - 1}} - \frac{i^2}{z^i}$

Thus, $(z - 1)S = 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}$

My question is how can I proceed? The numerator of each term is not 1 so I cannot use any formula to calculate the sum.

errorcodemonkey
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2 Answers2

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Let \begin{eqnarray*} S=\sum_{i=1}^{\infty} i^2 x^i. \end{eqnarray*} Now multiply by $(1-3x+3x^2-x^3)$ and note that for $i \geq 2$ \begin{eqnarray*} (i-3)^2-3(i-2)^2+3(i-1)^2-i^3=0. \end{eqnarray*} Examine the lower order terms more carefully, and we have \begin{eqnarray*} (1-3x+3x^2-x^3)S=x(1+x) \end{eqnarray*} giving the well known formula \begin{eqnarray*} S=\sum_{i=1}^{\infty} i^2 x^i =\frac{x(1+x)}{(1-x)^3}. \end{eqnarray*}

Donald Splutterwit
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  • Thanks for the help! How can you come up with multiply that magic number? In my experience, we usually multiply the original sum with a constant and then subtract it from the original sum. Is there a way we can find a constant? For example, the constant I came up with in the question is very intuitive. Also, my question is $\frac{i^2}{x^i}$ not $i^2x^i$ – errorcodemonkey Aug 28 '20 at 21:45
  • Multiplying by $(1-x)^3$ is perhaps not that magical when you know the final formula. Try doing $(1-x)^4$ for $\sum i^3 x^1$ and see https://en.wikipedia.org/wiki/Eulerian_number#Identities – Donald Splutterwit Aug 28 '20 at 21:53
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You have this $(z - 1)S = 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i} $ or, in summation notation, $(z - 1)S = \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \frac{i^2}{z^i} $.

We can now split this into sums we already know:

$\begin{array}\\ (z - 1)S &= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \dfrac{i^2}{z^i}\\ &= \sum_{k=0}^{i-1} \dfrac{2k}{z^k}+\sum_{k=0}^{i-1} \dfrac{1}{z^k}- \dfrac{i^2}{z^i}\\ &= 2\sum_{k=0}^{i-1} \dfrac{k}{z^k}+\sum_{k=0}^{i-1} \dfrac{1}{z^k}- \dfrac{i^2}{z^i}\\ \end{array} $

You can now plug in the known summations.

More generally, if

$S_m(z) =\sum_{k=0}^{\infty} \dfrac{k^m}{z^k} $, then $S_0(z) =\sum_{k=0}^{\infty} \dfrac{1}{z^k} =\dfrac{1}{1-1/z} =\dfrac{z}{z-1} $ and, for $m \ge 1$, $S_m(z) =\sum_{k=1}^{\infty} \dfrac{k^m}{z^k} $,

$zS_m(z) =\sum_{k=1}^{\infty} \dfrac{k^m}{z^{k-1}} =\sum_{k=0}^{\infty} \dfrac{(k+1)^m}{z^{k}} $ so

$\begin{array}\\ (z-1)S_m(z) &= zS_m(z)-S_m(z)\\ &=\sum_{k=0}^{\infty} \dfrac{(k+1)^m}{z^{k}}-\sum_{k=0}^{\infty} \dfrac{k^m}{z^k}\\ &=\sum_{k=0}^{\infty} \dfrac{(k+1)^m-k^m}{z^{k}}\\ &=\sum_{k=0}^{\infty} \dfrac{\sum_{j=0}^m \binom{m}{j}k^j-k^m}{z^{k}}\\ &=\sum_{k=0}^{\infty} \dfrac{\sum_{j=0}^{m-1} \binom{m}{j}k^j}{z^{k}}\\ &=\sum_{j=0}^{m-1} \binom{m}{j}\sum_{k=0}^{\infty} \dfrac{k^j}{z^{k}}\\ &=\sum_{j=0}^{m-1} \binom{m}{j}S_j(z)\\ \end{array} $

so $S_m(z) =\dfrac1{z-1}\sum_{j=0}^{m-1} \binom{m}{j}S_j(z) $ so that each $S_m(z) $ can be gotten in terms of the $S_j(z) $ for $j < m$.

In your case, the $2$ and the $1$ comes from $(k+1)^2-k^2 =2k+1 $.

marty cohen
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