Wikipedia states that the Euler-Maclaurin formula takes two versions. Here is the one that I am interested in: $$\sum_{k=m}^nf(k)=\int_m^nf(x)dx+\frac{f(n)+f(m)}2+\sum_{k=1}^{\lfloor\frac{p}2\rfloor}\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(n)-f^{(2k-1)}(m)\right)+R_p$$Let $m=0$ and $p=\infty$. Then we have $$\sum_{k=0}^n f(k)=\int_0^n f(x)dx+\frac{f(n)+f(0)}2+\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(n)-f^{(2k-1)}(0)\right)+R_\infty$$What are the required conditions for the sum and remainder to converge? Obviously, we want $f(x)$ to be continuous between $0$ and $n$ for the integral to be convergent, but I don't know about the other two. Recall that $$R_p=\frac{(-1)^{p+1}}{p!}\int_0^n f^{(p)}(x)P_p(x)dx$$Where $P_p(x)$ are the periodized Bernoulli polynomials. For the remainder to converge, we need $f^{(p)}(x)P_p(x)$ to approach zero (as $p$ approaches $\infty$). Otherwise, the remainder will keep oscillating. But I don't know all the functions that approach $0$ as they keep getting differentiated repeatedly (polynomials are one, but keep reading).
Additionally, I want the functions to have the condition that $f(x)$ approaches $0$ because I want it to make $$\sum_{k=0}^nf(k)=\sum_{k=0}^\infty(f(k)-f(k+n))$$converge (so I could take the derivative on this formula and the Euler-Maclaurin one then solve for $f(x)$).
Edit: As $p$ approaches infinity, the periodic Bernoulli functions become zero. This follows from the Fourier series expansion of it. If the variations in extrema of the $p$th derivative of $f$ grow slow enough, the remainder becomes zero. So we just need to find the functions that make the sum converge.
Edit 2: At first, I really wanted the sum with the Bernoulli coefficients to be included in the expressions. However, I realized that doing so doesn't achieve anything significant. If we let $p=1$ we have that $$\sum_{k=0}^nf(k)=\int_0^nf(x)+f'(x)(x-\lfloor x\rfloor-\frac12)dx+\frac{f(n)+f(0)}2$$