A Combinatorial Proof for the Power Sum

Probably the most common formula for the power sum \displaystyle \sum_{k=0}^n k^m is the one involving binomial coefficients and Bernoulli numbers B_k, sometimes called Faulhaber’s formula:

\displaystyle \sum_{k=0}^n k^m = \frac{1}{m+1} \sum_{k=0}^m \binom{m+1}{k} B_k n^{m+1-k}.

Historically, this, or a variant of it, was the first general formula for the power sum.

There are other ways to express the power sum, though.  The subject of this post is to give a combinatorial proof for the following one involving Stirling numbers:

\displaystyle \sum_{k=0}^n k^m = \sum_{k=1}^m \binom{n+1}{k+1} \left\{ m \atop k \right\} k! .

Proof: How many functions f are there from the set \{1, 2, \ldots, m+1 \} to the set \{1, 2, \ldots, n+1\} such that f(1) > f(j) for any j \in \{2, \ldots, m+1\}?

Answer 1: Condition on the value of f(1). If f(1) = k+1, then there are k choices for f(2), k choices for f(3), etc., for a total of k^m functions. Summing over all possible values of k gives the left side.

Answer 2: Condition on the number of elements in the range of f. (This is not the codomain, which is \{1, 2, \ldots, n+1\}, but the subset consisting of elements from the codomain that have something from \{1, 2, \ldots, m+1\} mapped to them.  The range may be the entire codomain or a proper subset of it.) If the range of f has k+1 elements, there are \binom{n+1}{k+1} ways to choose exactly which elements will comprise the range. The largest of these must be f(1), and the remaining m elements in the domain must be mapped to the other k elements in the range. There are \left\{ m \atop k \right\} ways to assign the remaining m elements to k nonempty groups, and then k! ways to assign these groups to the actual k remaining elements in the range. (More generally, the number of onto functions from an m-set to a k-set is \left\{ m \atop k \right\} k!.) Summing over all possible values of k gives the right side.

(I first gave this proof as an answer here on Math.SE.)

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This entry was posted in binomial coefficients, combinatorics, number theory, Stirling numbers. Bookmark the permalink.

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