Approximation of series using integral

Question

In notes of statistical physics I found the following approximation

$$\sum\limits_{n=0}^{\infty}F\left(n+\frac{1}{2}\right)\approx \int_{0}^{\infty}F(x)dx+\frac{1}{24}F'(0)$$

for $F$ such that the difference $F(n+1)-F(n)$ is "sufficiently small".

I don't have any idea how to prove it (if it is true) and what in this context means that above difference is "sufficiently small" ?

Answer 1

Actually, it's the mid-point rule for integrals

\begin{align*} \int_{a}^{b} f(x)\, dx &= (b-a)f\left( \frac{a+b}{2} \right)+ \frac{(b-a)^{2}}{24} [f'(b)-f'(a)] \\ &\quad \: - \frac{1}{6} \int_{a}^{\frac{a+b}{2}} (t-a)\left( t-\frac{a+b}{2} \right) \left( t+\frac{b-3a}{2} \right) f'''(t) \, dt \\ & \quad \: - \frac{1}{6} \int_{\frac{a+b}{2}}^{b} (t-b)\left( t-\frac{a+b}{2} \right) \left( t+\frac{a-3b}{2} \right) f'''(t) \, dt \\ &=(b-a)f\left( \frac{a+b}{2} \right)+ \frac{(b-a)^{2}}{24} [f'(b)-f'(a)]+O((b-a)^4) \\ \int_{0}^{N} F(x) \, dx & \approx \sum_{n=0}^{N} F\left( n+\frac{1}{2} \right)+ \frac{F'(N)-F'(0)}{24} \\ \end{align*}