I was trying to learn about finding the limit of an infinite series using Riemann sums and I derived the following conclusion using the basic Riemann definition of definite integration:
$$\int_{0}^{k}f(x)dx = \lim_{n\to \infty}\frac{k}{n}\sum_{r=0}^{n}f \left(\frac{rk}{n} \right)$$
However, my textbook seems to suggest that the integral can also be represented as follows:
$$\int_{0}^{k}f(x)dx = \lim_{n\to \infty}\frac{1}{n}\sum_{r=0}^{kn}f \left(\frac{r}{n} \right)$$
I tried to multiply and divide the latter expression by k but that’d need me to prove that
$$ \lim_{n\to \infty}\sum_{r=0}^{n}f \left(\frac{rk}{n} \right) = \lim_{n\to \infty}\frac{\displaystyle \sum_{r=0}^{kn}f \left(\frac{r}{n} \right)}{k}$$
I see the intuition behind this but am not sure if or how this could be true. Can this expression be proved by some means? If not, how can we say that the 2nd limit is equal to the definite integral from 0 to k?
Limit of Infinite Sum using Riemann sum
calculusdefinite integralsintegrationlimitsriemann sum
Best Answer
You can split the area under the curve into partitions $[x_0, x_1], ..., [x_{kn-1}, x_{kn}]$ such that where $x_0 = 0$, $x_{kn} = k$, and $\Delta x = \frac{1}{n}$. This will give the sum suggested by your textbook.