Generating function of $(-1)^n(n+1)^k$

Question

The power series $\displaystyle f_k(x)=\sum_{n=0}^\infty(-1)^n(n+1)^k x^n$ is of the form $$\frac{p_{k-1}(x)}{(1+x)^{k+1}}$$ where $p_{k-1}(x)$ is a degree $k-1$ polynomial satisfying the following conditions.

1. $p_0(x)=1$
2. $p_{k}(x)=p_{k-1}(x)+x(1+x)p_{k-1}'(x)-kxp_{k-1}(x)$

The second condition follows from termwise multiplication by $x$ and differentiation: $$f_{k+1}(x)=(xf_k(x))'.$$

The first few $p_k(x)$ are \begin{align*} p_1(x)&=1-x\\ p_2(x)&=1-4x+x^2\\ p_3(x)&=1-11x+11x^2-x^3\\ p_4(x)&=1-26x+66x^2-26x^3+x^4 \end{align*} One can see that the coefficients of $p_k(x)$ are alternating in sign and symmetric in magnitude.

Question: is there a closed-form formula for $p_k(x)$ (or its coefficients) for a general natural number $k$?

Answer 1

The question states

The power series $\displaystyle f_k(x)=\sum_{n=0}^\infty(-1)^n(n+1)^k x^n$ is of the form $$\frac{p_{k-1}(x)}{(1+x)^{k+1}}$$

The (unsigned) polynomial cofficients appear in the OEIS sequence A008292.

$\texttt{Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.}$

$\texttt{The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1))}$ $\texttt{for n >=1 lead directly to the triangle of Eulerian numbers.}$

$\texttt{1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1,}$

$\texttt{T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.}$ $\texttt{T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).}$

Use Mathematica to sum the infinite series as

Sum[(-z)^n*(n+1)^k, {n,0,Infinity}]
(* -(PolyLog[-k, -z]/z) *)
(* example for p_3(x) )
Sum[ (-z)^n(n+1)^(3+1), {n, 0, Infinity}]
(* (1 - 11z + 11z^2 - z^3)/(1 + z)^5 )
-PolyLog[-4, -z]/z //Simplify
( (1 - 11z + 11z^2 - z^3)/(1 + z)^5 *)

The general result is that $$ p_k(x) = -\text{Li}_{-k-1}( -x) $$ where $\,\text{Li}_n(x)\,$ is the polylogarithm function.

The answer to the question

Question: is there a closed-form formula for $p_k(x)$ (or its coefficients) for a general natural number $k$?

is "yes" for $p_k(x)$ if you consider the polylogarithm or Eulerian polynomials closed-form.

Alternatively, the finite summation for the coefficients

$$ \sum_{j=0}^n -(-1)^{n-j} (n-j)^k {{k+1}\choose j} $$

could be considered closed-form because it is a finite sum, but that is just a matter of opinion.

Answer 2

First,obtain the formula of: $$F(N,q) = \sum\limits_{{\rm{n = }}0}^{N - 1} {{q^n}} f(n),q \ne 0,1$$ Second,obtain the limit expression： $$L(q) = \mathop {\lim }\limits_{N \to \infty } F(N,q),0 < q < 1$$ then L(x) is the answer.Eulerian Polynomials A(M,t) is a special case of f(n)=n^M.
In this case: $$\eqalign{ & {p_k}(x) = \sum\limits_{g = 0}^{k + 1} {g!{S_2}(k + 2,g + 1)} {(1 + x)^{k + 1 - g}}{( - x)^g} \cr & = \sum\limits_{g = 0}^{k + 1} {{{( - 1)}^{k + 1 - g}}g!{S_2}(k + 1,g)} {(1 + x)^{k + 1 - g}} = \sum\limits_{g = 0}^{k + 1} {\left\langle {_g^k} \right\rangle } {( - x)^g} \cr} $$

$$\left\langle {_g^k} \right\rangle {\kern 1pt} {\kern 1pt} {\kern 1pt} is{\kern 1pt} {\kern 1pt} {\kern 1pt} Eulerian{\kern 1pt} {\kern 1pt} {\kern 1pt} Numbers,{S_2}(k,g){\kern 1pt} {\kern 1pt} is{\kern 1pt} {\kern 1pt} Stirling{\kern 1pt} {\kern 1pt} Number{\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} the{\kern 1pt} {\kern 1pt} {\kern 1pt} \sec ond{\kern 1pt} {\kern 1pt} {\kern 1pt} kind$$

Formal Calculations has various expressions and widespread generalizations of Eulerian polynomials.

Answer 3

I answered this as part of this question:

Does $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k =cx-x\ln(x)+x\ln(1-x) $ for $0 < x < 1$ for some real $c$? If so, what is $c$?

The other part was wrong.

Answer 4

Here we derive the coefficients of $p_k$ and show $p_k$ is a polynomial of degree $k$.

We obtain for $k\geq 1$: \begin{align*} \color{blue}{p_{k-1}(x)}&=(1+x)^{k+1}f_k(x)\\ &=\left(\sum_{m=0}^{k+1}\binom{k+1}{m}x^m\right)\left(\sum_{n=0}^{\infty}(-1)^n(n+1)^kx^n\right)\\ &=\sum_{q=0}^{\infty}\left(\sum_{{{m+n=q}\atop{m,n\geq 0}}\atop{m\leq k+1}} \binom{k+1}{m}(-1)^m(n+1)^k\right)x^q\tag{1}\\ &=\sum_{q=0}^{\infty}\left(\sum_{m=0}^{\min\{q,k+1\}}\binom{k+1}{m}(-1)^{q-m}(q-m+1)^k\right)x^q\tag{2}\\ &\,\,\color{blue}{=\sum_{q=0}^{k-1}\left(\sum_{m=0}^{q}\binom{k+1}{m}(-1)^{q-m}(q-m+1)^k\right)x^q}\tag{3}\\ \end{align*} Note that we claim in (3) that $p_{k-1}$ a polynomial of degree $\leq k-1$ which is not obvious and has to be shown.

When looking at the coefficient of $x^q, 0\leq q\leq k-1$ of $p_{k-1}$ \begin{align*} \color{blue}{[x^q]p_{k-1}(x)=\sum_{m=0}^{q}\binom{k+1}{m}(-1)^{q-m}(q-m+1)^k}\tag{4} \end{align*} we see the finite sum (4) is not in closed form. See for instance the reasoning of R. P. Stanley given in this answer.

Comment:

• In (1) we multiply out the series using Cauchy multiplication.

• In (2) we eliminate $n$ replacing it with $q-m$. We also take care when setting the upper index of the inner sum. We note that $\binom{k+1}{m}=0$ if $m>k+1$.

• In (3) we claim we have a polynomial of degree $\leq k-1$. In order to show it we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance \begin{align*} m^k=k![z^k]e^{mz} \end{align*} We claim \begin{align*} \color{blue}{\sum_{m=0}^q\binom{k+1}{m}(-1)^{q-m}(q-m+1)^k=0\qquad\qquad q\geq k}\tag{5} \end{align*}

We obtain for $q\geq k$: \begin{align*} \color{blue}{\sum_{m=0}^{k+1}}&\color{blue}{\binom{k+1}{m}(-1)^m(q-m+1)^k}\tag{6}\\ &=\sum_{m=0}^{k+1}\binom{k+1}{m}(-1)^mk![z^k]e^{(q-m+1)z}\tag{7}\\ &=k![z^k]e^{(q+1)z}\sum_{m=0}^{k+1}\binom{k+1}{m}(-1)^me^{-mz}\\ &=k![z^k]e^{(q+1)z}\left(1-e^z\right)^{k+1}\tag{8}\\ &=k![z^k]e^{(q+1)z}\left(-z-\frac{z^2}{2}-\cdots\right)^{k+1}\tag{9}\\ &\,\,\color{blue}{=0} \end{align*} and the claim (5) follows.

Comment:

• In (6) we set the upper limit of the sum to $k+1$ since $\binom{k+1}{m}=0$ if $q>k+1$. If $q=k$ we can also set the upper limit to $k+1$ since in this case we have $(k-m+1)^k=0$ if $m=k+1$.

• In (7) we apply (5) and use the linearity of the coefficient of operator in the next line.

• In (8) we apply the binomial theorem.

• In (9) we expand the exponential function noting that it contains exponents of $z$ always greater than $k$. Since $e^{(q+1)z}=1+(q+1)z+\cdots$ contains exponents of $z$ greater or equal to zero, the coefficient of $z^k$ in (9) is zero.