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.
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.
You should replace the infinite sum with a finite sum with remainder, see Wikipedia, in order to decide your question.
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.