I saw that a similar question had been asked, but mine is specifically about what is wrong with this sum representation.
So, if we were to imagine a circle of radius 1, that radius can be divided into many small divisions of a value we can call n, we can represent the divisions like so:
$$\frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \frac{4}{n} ... (\frac{n}{n}=1).$$
These divisions can be multiplied by $2\pi$ (from $2\pi*r$) to represent a circumference within the circle of radius 1. These circumferences would be represented like so:
$$\frac{1*2\pi}{n}, \frac{2*2\pi}{n}, \frac{3*2\pi}{n}, \frac{4*2\pi}{n} ... \frac{n*2\pi}{n}=1.$$
$$2\pi(\frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \frac{4}{n} ... (\frac{n}{n}=1)).$$
As $n\to \infty$, we can see that the divisions become infinitely small and the circumferences make up the interior of the circle. The sum of these circumferences should be the area of the circle. I thought this could be represented as follows:
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k}{n}*2\pi,$$
leading to a value of 1, but this sum goes to infinity. Where is the reasoning flawed? (Also not sure whether or not I should have subtracted the outer circumference of $2\pi$ from the sum.)
You cannot add circumferences to get an area, but you can multiply each circumference by a small thickness (and this thickness decreases with $n$) to get an area of a ring, and sum these areas. One such approach produces $$\sum_{k=1}^n \left(2 \pi \frac{k}{n} \cdot \frac{1}{n}\right) = 2 \pi \frac{n+1}{2n} \to \pi$$ which can also be viewed as a Riemann sum for $$\int_0^1 2 \pi r \, dr = \pi.$$