Is my use of derivatives correct in this infinite sum?

Question

I was inspired by a question that seeked to find a generating function for the sum of powers.
We set $s(n,p)=\sum_{k=0}^n k^p$, now we are looking for $G(x,p)= \sum_{n=0}^{\infty}s(n,p)x^n$.
Consider $\frac{\partial^p}{\partial y^p} e^{ky}|_0=k^p$ and sum over $k$ to $n$ to obtain. $$\sum_{k=0}^n k^p=\sum_{k=0}^n \frac{\partial^p}{\partial y^p} e^{ky}|_0=\frac{\partial^p}{\partial y^p} \sum_{k=0}^ne^{ky}|_0=\frac{\partial^p}{\partial y^p} \frac {e^{(n+1)y}-1}{e^y-1}|_0=s(n,p)$$ Now plug into $G(x,p)$ to get: $$\begin{align}\sum_{n=0}^{\infty}\frac{\partial^p}{\partial y^p} \frac {e^{(n+1)y}-1}{e^y-1}|_0x^n & =\frac{\partial^p}{\partial y^p} \sum_{n=0}^{\infty}\frac {e^{(n+1)y}-1}{e^y-1}x^n|_0 \\ &= \frac{\partial^p}{\partial y^p} \left((e^y-1)^{-1}\left(\sum_{n=0}^{\infty}e^{(n+1)y}x^n-\frac {1}{1-x}\right)\right)|_0 \\ & =\frac{\partial^p}{\partial y^p} \left((e^y-1)^{-1}\left(\frac {e^y}{1-xe^y}-\frac {1}{1-x}\right)\right)|_0\\ & = \frac{\partial^p}{\partial y^p} \frac{1}{(1-xe^y)(1-x)}|_0\end{align}$$ The general expression for the derivative is:$$\frac{\partial^p}{\partial y^p} \frac{1}{(1-xe^y)}=\frac{xe^yA_p(xe^y)}{(1-xe^y)^{p+1}}$$ Where $A_p(x)$ is the $p$th Eulerian Polynomial. This gives an expression for $G(x,p)$, namely: $$G(x,p)=\frac {xA_p(x)}{(1-x)^{2+p}}$$ Is this correct? Can I justify pulling the partial out of the sum, or pulling the $x$ terms into the partial and then the partial out of the sum?
This process seems natural to me, but I am not sure whether it is rigorous, where would rigor problems arise?

Answer 1

All calculations are valid.

• Formal power series: Here we consider the integral domain of formal power series $F[[x]]$ with coefficients from an underlying field $\mathcal{F}$ (typically $\mathbb{R}$ or $\mathbb{C}$).

  The elements $f\in F[[x]]$ are purely algebraic objects and can be considered as sequence $f=(f_n)_{n\geq 0}$ which are written for convenience only as series \begin{align*} f=\sum_{n=0}^\infty f_n x^n \end{align*} No value is assigned to the symbol $x$, so $f$ should not be regarded as an ordinary infinite series. There is no norm imposed on $\mathcal{F}$ and so we cannot speak of convergence, even if we replace $x$ with an element of $\mathcal{F}$.

  Evaluation at $x=0$ is somewhat special, since there is not an infinite number of terms involved to obtain the resulting value. It can be simply regarded as the mapping of the sequence $f=\{f_0,f_1,f_2,\ldots\}$ to $\{f_0,0,0,\ldots\}$.

• Differentiation: The differentiation operator $D_x:=\frac{d}{dx}$ is treated purely algebraically as an operator acting on the sequence $f=\{f_0,f_1,f_2,\ldots\}$. It produces the sequence $D_xf:=\{f_1,2f_2,3f_3,\ldots\}$. Since this operator has many properties of differentation, familiar from analysis we often write $f^\prime:=D_xf$ and call $f^\prime$ the derivative of $f$. Thus, if \begin{align*} f(x)=f_0+f_1x+f_2x^2+\cdots \end{align*} then \begin{align*} f^\prime(x)=f_1+2f_2x+3f_3x^2+\cdots \end{align*}

• Evaluation: If we want to evaluate a formal power series $f$ at a specific value, we have to consider a framework where the wanted behaviour of convergence, norm, completeness, multiplication of power series, etc. is available. For this reason we typically consider a Banach Algebra $\mathcal{A}$.

  If $\mathcal{A}$ is a Banach Algebra and $f(x)=\sum_{n=0}^\infty f_nx^n$ is a formal power series, we consider the finite partial sums \begin{align*} f_N(x)=\sum_{n=0}^N f_nx^n \end{align*} and it is possible to consider $f_N(x)$ for any element $x\in \mathcal{A}$. We can then ask if $\left(f_N\right)_{N\geq 0}$ is a Cauchy sequence. If it is, then its limit is an element of $\mathcal{A}$.

  On the set of those $x\in\mathcal{A}$ for which the formal power series converges it defines a function with values in $\mathcal{A}$.

Hint: A wonderful classic which extensively describes formal power series is volume 1 of Applied and Computational Complex Analysis by P. Henrici.