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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.)

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The problem is that your functions don't have $a_n(0)=a_n(1)$ for $n>1$. The proof you linked to uses the fact that $B_1(0)=-B_1(1)$ and that $B_n(0)=B_n(1)$ for $n>1$; this is why the sum of the function values appears in the integration by parts for $n=1$ but only boundary terms appear in the further integrations by parts for $n>1$.

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    $\begingroup$ Note that $B_n(1) - B_n(0) = \int_0^1 B'_n(x) \, \mathrm{d}x = n\int_0^1 B_{n-1}(x) \, \mathrm{d}x$. $\endgroup$ Commented May 18, 2011 at 2:50

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