Originally Posted by

**felper** Note that the sum you have, is a Riemann sum of the form:

$\displaystyle \dfrac{b-a}{n}\displaystyle\sum_{i=0}^nf\left(a+i\dfrac{b-a}{n}\right)$

Where $\displaystyle \dfrac{b-a}{n}$ is a partition of the interval, $\displaystyle a+i\dfrac{b-a}{n}$ is a selection of points of each interval defined by the partition. The norm of the partition is regular, that means that every interval defined by the partition has the same lenght. This Riemann sum is defined by a aritmetic progression. In more abstract sence, a Riemann sum just have to look like this:

$\displaystyle \sum_{i=0}^nf(a_i)\Delta x_i$

The Riemann integration tell us that if a Riemann sum converges when the norm of the partition tends to 0, the function is Riemann-integrable. So, can you tell me which is the partition in this particular case? can you express your sum like a Riemann sum?