Limit of series involving sum of cosines $\lim_{n \to \infty} \sum_{k=0}^{n} \cos(\frac{2k\pi}{n})$ How to approach finding the limit of series $\lim_{n \to \infty}S_n $ where $ S_n = \sum_{k=0}^{n} \cos(\frac{2k\pi}{n})$? Computer simulations show that is tends to $1$ which is the first term but I have a difficult time wrapping my head around it. The problem is home grown, I was working on approximating a volume of a solid where this came up. Any hints are appreciated.
 A: Consider a unit vector initially horizontal that rotates $n$ times by $\dfrac{2\pi}n$ and is added to itself. The vector will describe a regular $n$-gon, which is closed. By projecting on the coordinate axis, we obtain
$$\sum_{k=0}^{n-1}\cos\frac{2k\pi}n=\sum_{k=0}^{n-1}\sin\frac{2k\pi}n=0.$$

A: Substitute
$$
\cos\frac{2k\pi}{n}=\frac{e^{j\frac{2k\pi}{n}}+e^{-j\frac{2k\pi}{n}}}{2}
$$
inside the $\sum$ argument and use geometric series before finding the limit.
A: Your sums are always exactly equal to $1$.
Using $\cos t = \Re \left(e^{it}\right)$ for $t \in \mathbb{R}$ and noting that for $k=0,\ldots,n-1$:
$$z_k := e^{i\frac{2k\pi}{n}} \text{ are the zeros of } z^n -1 = 0$$
Vieta gives
$$\sum_{k=0}^{\color{blue}{n-1}}z_k = 0$$
Hence,
$$\sum_{k=0}^{\color{blue}{n}}z_k = e^{2\pi i}=1$$
A: Hint 1: Use the cosine with angles in AP formula. You can derive it using the telescopic method, otherwise you can easily look it up.
Hint 2: Aliter: Define a sequence $S+iT$ where $T$ gives the summation of the sine counterparts of $S$. Then use GP serie, and finally take the real part of the result.
