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In the Euler–Maclaurin formula:

$$\sum_{n=a}^b \sim \int_a^b f(x)\;dx+\frac{f(a)+f(b)}{2}+\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))$$

can I neglet the series with Bernoulli numbers? Thanks, Anna.

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  • $\begingroup$ i.e. can I: $ \int_{a}^{b}f(x)dx+\frac{f(a)+f(b)}{2}+\sum_{k=1}^{\infty}\frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))\sim \int_{a}^{b}f(x)dx+\frac{f(a)+f(b)}{2}$? $\endgroup$
    – Anna
    Feb 29 '12 at 11:06
  • $\begingroup$ Unless you give us more information about the shape of (the derivatives of) $f$ that's impossible to answer. $\endgroup$
    – TMM
    Feb 29 '12 at 11:32
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You should replace the infinite sum with a finite sum with remainder, see Wikipedia, in order to decide your question.

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Let $f(x)=e^x$. Then $$\sum_0^ne^m=(e^{n+1}-1)/(e-1)$$ is asymptotic to $Ce^n$ with $C=e/(e-1)$, $$\int_0^ne^x\,dx+(1/2)(e^0+e^n)=e^n-1+(1/2)e^n+(1/2)$$ is asymptotic to $(3/2)e^n$. $C\ne3/2$, so the sum on the left is not necessarily asymptotic to the first two terms on the right.

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