Faulhaber's Formula - Proof

Proof

Let the sums S_p(n) be defined as

It follows from this that S_p(0+1)=S_p(0)+(0+1)p and so S_p(0)=0.

If the closed forms for S_p(n) can be expressed as polynomials they will not have constant terms, and will be of the form

for some matrix of coefficients a_p,k. Substituting this into the inductive part of the definition above, applying the binomial theorem, and rearranging, we have

The double sum on the left hand side can be rearranged, noting that j≤k

Aligning coefficients then gives the relation

If p is a natural number (the reasoning so far has not assumed it is), then the right hand side is 0 when j>p, and we can assert that a_p,k=0 for all k>p. Note that this is not the only possible solution; an infinite series that satisfied this property for its coefficients would also be valid, and would correspond to a closed form with a periodic component that was 0 at every natural number, such as sin(πn). The multiple solutions are an immediate result of the closed forms being defined only at the integers.

Asserting now that the matrix of coefficients is triangular, and multiplying both sides by j!, we have the recurrence relation giving all coefficients for any p (the first at j=p, the next at j=p-1, etc.).

which makes use of the Pochhammer symbol for the falling factorial.

Just as there is a diagonal relationship between binomial coefficients and between falling factorials, there is such a relationship between these coefficients, which can be demonstrated by 'breaking' the falling factorials at an arbitrary number t≤j.

Substituting k=k'+t and rearranging gives

This is of the same form as the original recurrence relation, which shows that

In particular, for k'=0, we have

The linear coefficients a_p,0 are thus fundamental. These are the Bernoulli numbers, and so we have the final form

with each Bernoulli number derivable from this formula itself at n=1

and so B₁=1/2 here.