When can the sum of a series be calculated using an integral

Question

In some cases, we can think of a series as a Riemann sum, and then calculate it using an integral of a certain function. E.g. (Problems in Mathematical Analysis III: Integration, Kaczor & Nowak, p.7) $$ \lim_{n\to \infty} \sum_{i=1}^{2n} \frac{1}{n+i} = \lim_{n\to \infty} \frac{1}{2n} \sum_{i=1}^{2n} \frac{1}{\frac{1}{2} + \frac{i}{2n}} = \int_0^1 \frac{1}{\frac{1}{2} + x} \,dx = \ln(3).$$

My question is: what are the necessary and sufficient conditions on a series, to be calculable in this manner.

Clarification: what are the necessary and sufficient conditions on a series, so as to be manipulable into a Riemann sum ?

Answer 1

You can do it when you can rewrite your sum as a Riemann sum:

$$\lim_{n \rightarrow +\infty} \frac{b-a}{n}\sum_{i=1}^n f(\xi_i) = \int_a^b f(x)dx$$ where $\xi_i$ is a point in the $i$-th sub-interval of the partition of $[a,b]$.

Answer 2

Thanks to Dr. Riemann, you have a simple test. Take the first $n$ of the series sequentially and out of these terms, if you can express the $r$-th of the series in the form of $f(r/n)$ where $f$ is some function integrable in $(0,1)$ then making $n \to \infty$, you can evaluate the sum of the series using the integral $$ sum = \int_{0}^1 f(x)dx $$