Faulhaber's Theorem on Power Sums

In a recent work, Zielinski used Faulhaber's formula to explain why the odd Bernoulli numbers are equal to zero. Here, we assume that the odd Bernoulli numbers are equal to zero to explain Faulhaber's formula.

The problem of finding the sum of a polynomial's values is considered. In particular, for any $n\geq 3$, the explicit formula for the sum of the $n$th powers of natural numbers $S_n=\sum_{x=1}^{m}x^{n}$ is proved: $$\sum_{x=1}^{m}x^{n}=(-1)^{n}m(m+1)(-\frac{1}{2}+\sum_{i=2}^{n}a_i(m+2)(m+3)...(m+i)),$$ here $a_i=\frac{1}{i+1}\sum_{k=1}^{i}\frac{(-1)^{k}k^{n}}{k!(i-k)!}$, $(i=2,3,...,n-1)$, $a_n=\frac{(-1)^n}{n+1}$. Note that this formula does not contain Bernoulli numbers.

In this article we present a simple proof of Borevich-Shafarevich's method to compute the sum of the first n natural numbers of the same power. We also prove several properties of Bernoulli's numbers.

What is a general expression for the sum of the first n integers, each raised to the mth power, where m is a positive integer? Answering this question will be the aim of the paper....We will take the unorthodox approach of presenting the material from the point of view of someone who is trying to solve the problem himself. Keywords: analogy, Johann Faulhaber, finite sums, heuristics, inductive reasoning, number theory, George Polya, problem solving, teaching of mathematics

For integer $k \geq 0$, let $S_k$ denote the sum of the $k$th powers of the first $n$ positive integers $1^k + 2^k + \cdots + n^k$. For any given $k$, the power sum $S_k$ can in principle be determined by differentiating $k$ times (with respect to $x$) the associated exponential generating function $\sum_{k=0}^{\infty}S_k x^k/k!$, and then taking the limit of the resulting differentiated function as $x$ approaches $0$. In this paper, we exploit this method to establish a coup...

In this paper we present a generalization of Faulhaber's formula to sums of arbitrary complex powers $m\in\mathbb{C}$. These summation formulas for sums of the form $\sum_{k=1}^{\lfloor x\rfloor}k^{m}$ and $\sum_{k=1}^{n}k^{m}$, where $x\in\mathbb{R}^{+}$ and $n\in\mathbb{N}$, are based on a series acceleration involving Stirling numbers of the first kind. While it is well-known that the corresponding expressions obtained from the Euler-Maclaurin summation formula diverge, ou...

We realize that geometric polynomials and p-Bernoulli polynomials and numbers are closely related with an integral representation. Therefore, using geometric polynomials, we extend some properties of Bernoulli polynomials and numbers such as recurrence relations, telescopic formula and Raabe's formula to p-Bernoulli polynomials and numbers. In particular cases of these results, we establish some new results for Bernoulli polynomials and numbers. Moreover, we evaluate a Faulha...

In this paper, we will define general Eulerian numbers and Eulerian polynomials based on general arithmetic progressions. Under the new definitions, we have been successful in extending several well-known properties of traditional Eulerian numbers and polynomials to the general Eulerian polynomials and numbers.

In this paper, we derive a formula for the sums of powers of the first $n$ positive integers, $S_k(n)$, that involves the hyperharmonic numbers and the Stirling numbers of the second kind. Then, using an explicit representation for the hyperharmonic numbers, we generalize this formula to the sums of powers of an arbitrary arithmetic progression. Moreover, we express the Bernoulli polynomials in terms of hyperharmonic polynomials and Stirling numbers of the second kind. Finall...

In this paper we study recurrences concerning the combinatorial sum $[n,r]_m=\sum_{k\equiv r (mod m)}\binom {n}{k}$ and the alternate sum $\sum_{k\equiv r (mod m)}(-1)^{(k-r)/m}\binom{n}{k}$, where m>0, $n\ge 0$ and r are integers. For example, we show that if $n\ge m-1$ then $$\sum_{i=0}^{\lfloor(m-1)/2\rfloor}(-1)^i\binom{m-1-i}i [n-2i,r-i]_m=2^{n-m+1}.$$ We also apply such results to investigate Bernoulli and Euler polynomials. Our approach depends heavily on an identity e...