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Feb 6, 2022 at 19:23 comment added François Brunault One should mention here that for many usual functions $f$ and for fixed $m,n$, the error term $R_p$ does not converge to 0 when $p \to \infty$. Euler-Maclaurin provides an asymptotic expansion, but it does not always converge.
Dec 8, 2021 at 18:05 history edited LSpice CC BY-SA 4.0
Period at the end
Dec 8, 2021 at 17:01 answer added juan timeline score: 1
Dec 8, 2021 at 15:17 history edited Max Lonysa Muller CC BY-SA 4.0
Added a note to clarify the question
Dec 7, 2021 at 21:05 comment added Max Lonysa Muller I would also consider 'closed-form' a synonym of 'exact' in this case: en.wikipedia.org/wiki/Closed-form_expression
Dec 7, 2021 at 20:59 comment added Max Lonysa Muller (In the comment above, I meant $p \to \infty$ instead of $n \to \infty$ )
Dec 7, 2021 at 20:47 comment added Max Lonysa Muller (ii) It becomes exact in the sense that it isn't an approximation, and yields the correct answer for any sum of powers. (iii) I mean any convergent infinite series. I'm not looking for a specific one, but for any evaluation of an infinite series by means of the Euler-Maclaurin summation formula. As for an evaluation of the terms (there are four in the formula, and three if one can show that $R_{p}$ tends to zero for $n \to \infty$), please refer to my answer to (i)
Dec 7, 2021 at 20:41 comment added Max Lonysa Muller @IosifPinelis (i) Please note I said it usually can't be evaluated exactly. By an exact evaluation, I mean something like $\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6} $ and many others, like the ones listed over here: en.wikipedia.org/wiki/Series_(mathematics)
Dec 7, 2021 at 20:36 comment added Iosif Pinelis (iii) "I wonder whether there are infinite series for which all terms of the formula above can be evaluated." What infinite series? (You only have a finite sum in your post.) Also, what do you mean, exactly, by "all terms of the formula" and, again, by "can be evaluated"?
Dec 7, 2021 at 20:35 comment added Iosif Pinelis (i) "the third term (the sum involving the Bernoulli numbers) and the remainder term cannot be evaluated exactly". Why (in what sense) can't the sum involving the Bernoulli numbers be evaluated exactly? (ii) "If the higher derivatives eventually become zero at the start and end points, the formula becomes exact." Exact in what sense, exactly?
Dec 7, 2021 at 20:01 history asked Max Lonysa Muller CC BY-SA 4.0