Non-constructive proof that $\sum_{j=1}^n j^k$ is a polynomial $p(n)$ of degree $k+1$

Question

So it can be shown that there are special polynomials (I forget their name) $p_k$ of degree $k$ that satisfy $\sum_{j=1}^n p_k(j) = n^{k+1}$, and that these polynomials are linearly independent so that a sum of any polynomial from $j=1$ to $n$ is equal to a polynomial in $n$ of one degree higher. However I was wondering if there's a non-constructive way to prove this. For example, if we integrate $\int_{0}^n x^k dx$ we get $n^{k+1}/(k+1)$ so we know asymptotically the summation grows like a polynomial. Is there any way to turn this into a proof, e.g. assume there is an entire function that has the asymptotic growth rate and takes on the prescribed values and then show that the entire function must be a polynomial?

Answer 1

Let $$S(n,k)=\sum_{j=1}^{n}j^{k}.$$ Then notice that $\frac{1}{k!}S(n,k)$ is the coefficient of $x^{k}$ in the power series expansion around $x=0$ of $$e^{x}+e^{2x}+e^{3x}+\cdots+e^{nx}.$$ That is $$\frac{e^{(n+1)x}-1}{e^{x}-1}=\sum_{k\geq0}\frac{S(n,k)}{k!}x^{k}.$$ Using this expansion, we may define a function a function $S(t,k)$ for $t\in\mathbb{R}$, and we will show that $S(t,k)$ is a polynomial of degree $k+1$. Lets look at the function $$f_t(x)=e^{x(t+1)}/(e^{x}-1).$$ The $j^{th}$ derivative with respect to $t$ equals $$\frac{d^j}{dt^j} f_t(x)=\frac{x^je^{x(t+1)}}{e^x-1}.$$ For fixed $t$, since $$\lim_{x\rightarrow0}\frac{x^{r}e^{x(t+1)}}{e^{x}-1}=\begin{cases} 1 & \text{if}\ r=1\\ 0 & \text{if}\ r>1 \end{cases}$$ we see that by expanding this as a power series in $x$ around $0$, the coefficients of $1,x,\dots,x^{j-2}$ will all equal $0$, and the coefficient of $x^{j-1}$ will be non-zero. Thus, by switching the order of differentiation and summation, it follows that $$\frac{d^{k+2}}{dt^{k+2}}S(t,k)= 0$$ and $$\frac{d^{k+1}}{dt^{k+1}}S(t,k)= 1.$$ Hence we conclude that $S(n,k)$ is a polynomial of degree $k+1$.

Answer 2

Consider the function $f_k(n)=\sum_{j=1}^n j^k$. Let $\Delta$ be the difference operator, so for a function $f$ we have $(\Delta f) (n) = f(n+1)-f(n)$. Observe that $(\Delta f_k)(n) = (n+1)^k$, a polynomial of order $k$ in $n$. Now $(\Delta^2 f_k)(n) = (n+2)^k -(n+1)^k$, and the top terms in $n$ cancel, so this is a polynomial of order $k-1$ in $n$. Each further time we apply $\Delta$ we drop the degree by one, so $\Delta^{k+2} f_k = 0$ (and in fact $\Delta^{k+1} f_k \neq 0$).

Now I claim that any function of $n$ killed by $\Delta^r$ but not $\Delta^{r-1}$ is a polynomial of degree $r-1$ (this is clear for $r=1$). The proof is by induction: suppose $\Delta^r f \neq 0 = \Delta^{r+1}f$, so that $\Delta^r f = C$, a constant. Observe that $\Delta^m n^m = m!$. Let $g(n) = f(n)-Cn^r/r!$. We have $\Delta^r g = 0$, and so by induction $g$ is polynomial of degree $r-1$, and therefore $f$ is polynomial of degree $r$.

This is the discrete analogue of the theorem that if the $r$th derivative of a function is the zero function then that function is polynomial of degree at most $r-1$.

Answer 3

Let $V = \mathbf{Q}[x]$, and let $V_r$ be the subspace of polynomials of degree $\leq r$.

Define a mapping $\Delta \colon V \to V$ by $\Delta(p)(x) = p(x) - p(x-1)$. Then:

1. It is clear that $\Delta$ is a linear mapping.

2. $\Delta(V_r) \subseteq V_{r-1}$.
   Proof. Let $p(x)$ have degree $s \leq r$. Then $p(x)$ and $p(x-1)$ have the same leading term, of degree $s$. Therefore $\Delta(p)$ has degree $\leq s - 1 \leq r - 1$.

3. $\ker \Delta$ consists of the constant polynomials, hence its dimension is $1$.
   Proof. Let $p \in \ker \Delta$. Then by induction, $p(a)$ has the same value $k$ for all natural numbers $a$. The polynomial $p(x) - k$ has infinitely many roots, hence is zero. Therefore $p$ is constant. The converse is obvious.

Now consider $\Delta$ as a linear mapping from $V_{k+1}$ to $V_k$ (by points 1 and 2). Its kernel has dimension $1$ by point 3. Comparing dimensions, we see that we must have $\Delta(V_{k+1}) = V_k$. Thus there is some polynomial $p$ of degree at most $k + 1$ for which $\Delta(p) = x^k$. Moreover, applying point 2 for $r = k$, we see that $p$ must have degree exactly $k + 1$.

Thus we have $$\sum_{j = 1}^n j^k = \sum_{j = 1}^n [P(j) - P(j-1)] = P(n) - P(0),$$ completing the proof.