It may be that I have not picked up the proof, but I cannot see where the third condition of Bernoulli polynomials, given below, is used in the derivation of the Euler-Maclaurin summation formula.
The Bernoulli polynomials are defined inductively by $ b_0(x) = 1, $ b_n'(x) = nb_{n-1}(x) \:\text{ and }\: \int_0^1 b_n(x) = 0 \:\text{ for }\: n \geq 1.
Suppose instead that we defined the alternative Bernoulli polynomials as follows, $ a_0(x) = 1, $ a_n'(x) = na_{n-1}(x) \:\text{ and }\: a_n(0) = 0 \:\text{ for }\: n \geq 1. so that the new third condition essentially gives $a_n(x) = x^n$ for all $n$, thus simplifying matters. Then the periodic Bernoulli functions used in the proof would be $A_n(x) = a_n(\{x\})$, where $\{x\}$ denotes the fractional part of $x$.
Using these functions, can anyone show me where the proof of Euler-Maclaurin breaks down? (Assume we are using the elementary proof by induction that is explained in detail on Wikipedia, for example.)