Sum of complex e's

Note that $e^{\frac {2\pi}ni}$ is the $n$ th root of unity, $\omega$ . Then

$\begin{aligned} \sum_{j=1}^n e^{\frac {2\pi i}nj} & = \omega + \omega^2 + \omega^3 + \cdots + \blue{\omega^n} & \small \blue{\text{Note that }\omega^n = 1} \\ & = \omega + \omega^2 + \omega^3 + \cdots + \blue 1 \\ & = 1 + \omega + \omega^2 + \cdots + \omega^{n-1} \\ & = 0 & \small \blue{\text{Another property of }n\text{th roots of unity.}} \end{aligned}$

Since $\displaystyle \sum_{j=1}^n e^{\frac {2\pi i}nj} = 0$ for all $n \ge 2$ , $\displaystyle \lim_{n \to \infty} \sum_{j=1}^n e^{\frac {2\pi i}nj} = \boxed 0$ .

The limit is

$\displaystyle \lim_{n\to \infty } \dfrac {e^{\frac{2πi}{n}}(e^{2πi}-1)}{e^{\frac{2πi}{n}}-1}$

$=\boxed 0$ .

The $n$ points in the complex plane given by $e^{2\pi i k/n}$ for $k=1,2,\cdots,n$ form the vertices of a regular $n$ -gon centred at $0$ , so this sum is equal to $\boxed0$ for all $n$ .