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Here I have $x\in\mathbb{R}_+$ and $x < 1$.

I would like to evaluate the following sum:

$$\sum_{i=0}^\infty i^nx^i.$$

I know that $$\sum_{i=0}^\infty x^i=\dfrac{1}{1-x}.$$

So I started calculating the derivative of the above formula. I found something like:

$$\sum_{i=0}^\infty i^nx^i=\dfrac{P_n(x)}{(1-x)^{n+1}},$$ where $P_n(x)$ is a polynomial of degree $n$.

$$P_0(x)=1,$$ $$P_1(x)=x,$$ $$P_2(x)=x+x^2,$$ $$P_3(x)=x+4x^2+x^3,$$ $$P_4(x)=x+11x^2+11x^3+x^4,$$ $$P_4(x)=x+26x^2+66x^3+26x^4+x^5,$$ $$\cdots.$$

Is there a general formula for $P_n(x)$?

Kira
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2 Answers2

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The coefficients of $P_n(x)$ are called the Eulerian numbers and there is a corresponding Euler's triangle. They satisfy $$ \genfrac{\langle}{\rangle}{0}{}{n}{k} = (k+1)\genfrac{\langle}{\rangle}{0}{}{n-1}{k}+(n-k)\genfrac{\langle}{\rangle}{0}{}{n-1}{k-1}, $$ for integer $n>0$ as well as $$ \genfrac{\langle}{\rangle}{0}{}{n}{k}=\genfrac{\langle}{\rangle}{0}{}{n}{n-1-k} $$

Chappers
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kodlu
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Suppose $p_n(x) =\sum_{i=0}^\infty i^nx^i $. Then $p_n'(x) =\sum_{i=0}^\infty i^{n+1}x^{i-1} $ so $x p_n'(x) =\sum_{i=0}^\infty i^{n+1}x^i =p_{n+1}(x) $.

Since $p_0(x) =\frac1{1-x} $, $p_1(x) =\frac{x}{(1-x)^2} $.

This leads to the conjecture that $p_n(x) =\frac{q_n(x)}{(1-x)^{n+1}} $ for some polynomial $q_n$.

$\begin{array}\\ p_n'(x) &=\left(\frac{q_n(x)}{(1-x)^{n+1}}\right)'\\ &=\frac{q_n'(x)(1-x)^{n+1}-q_n(x)(n+1)(1-x)^{n}}{(1-x)^{2n+2}}\\ &=\frac{q_n'(x)(1-x)-q_n(x)(n+1)}{(1-x)^{n+2}}\\ \end{array} $

so that $q_{n+1}(x) =x p_n'(x)(1-x)^{n+1} =x(q_n'(x)(1-x)-q_n(x)(n+1)) $ and this is a polynomial of degree one higher than $q_n(x)$.

The next step is to see how the coefficients of $q_{n+1}$ are gotten from those of $q_n$. When you do this. you will get the recurrence that kodlu has in his answer.

As in many of my answers, nothing original here.

marty cohen
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