The Euler-Maclaurin summation formula is \begin{eqnarray} \sum_{k = a}^{b} f(k) = \int_{a}^{b} f(t) \, dt + B_1 (f(a) + f(b)) + \sum_{n = 1}^{N} \frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N}, \end{eqnarray} where $B_{n}$ is the $n^{\text{th}}$-Bernoulli number taking $B_{1} = \tfrac{1}{2}$, and the remainder term is bounded by the following \begin{align} |R_{N}| \leq \frac{|B_{2N} |}{(2n)!} \int_{a}^{b} | f^{(2N)}(t) | \, dt. \end{align} for any arbitrary positive integer $N$. Is there a similar formula for nested sums of the form, \begin{eqnarray} \sum_{k_1 = a_1}^{b_1} \cdots \sum_{k_n = a_n}^{b_n} f(k_1, \dots, k_n). \end{eqnarray}



Yes! There's a whole chapter about it in this book.

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    $\begingroup$ The formula there (say in Theorem 2 there) is stated only for the case when $f$ is an exponential function. In fact, the Todd operator (in terms of which the formula there is stated), seems to be only defined on functions $f$ of the form $f(z)=\sum_{n=0}^\infty c_n^n z^n/n!$ with bounded $c_n$'s -- so that such functions are somewhat like the exponential functions.Strangely enough, the domain of the operator is not specified there. $\endgroup$ – Iosif Pinelis May 15 '17 at 4:27
  • $\begingroup$ I meant Theorem 10.2 there in the book. $\endgroup$ – Iosif Pinelis May 15 '17 at 14:52

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