Originally Posted by
NonCommAlg you should know that the convergence of$\displaystyle \int^{\infty} f(x) \ dx$ doesn't necessarily imply that $\displaystyle \sum^{\infty} f(n)$ is convergent. for example $\displaystyle \int_0^{\infty} \sin x^2 \ dx=\sqrt{\frac{\pi}{8}}$ but the series $\displaystyle \sum_{n=0}^{\infty} \sin n^2$ is divergent.