Showing that the sum of a complex power series is equal to an indicator variable

Question

When I evaluate this for some arbitrary values manually, I get the expected result of 1 when $\nu/n$ is an integer and 0 when it is not. However, trying to find a closed term for the sum is giving me trouble. I've tried rewriting in terms of cos and sin, simplifying, etc. I feel like it has something to do with the $e^{2i\pi}$ being able to simplify to 1 causing issues, because the final term I get using the geometric series is undefined when $\nu/n$ is an integer, but 0 when it is not. How should I go about finishing up this?

Answer 1

If $\frac{v}{n}$ is an integer then every term in the sum is $1$. Otherwise $$ \sum_{k=1}^ne^{\frac{2\pi ikv}{n}}=e^{\frac{2\pi iv}{n}}\sum_{k=0}^{n-1}e^{\frac{2\pi ikv}{n}}=e^{\frac{2\pi iv}{n}}\frac{1-e^{2\pi iv}}{1-e^{\frac{2\pi iv}{n}}}=0$$

Therefore $$ \frac{1}{n}\sum_{k=1}^ne^{\frac{2\pi ikv}{n}}=1_{\frac{v}{n}\in\mathbb{Z}}$$ as claimed.