Loser66 one year ago Need help understanding the concept Let $$z = e^{i\theta} \in S^1$$ , $$S^1 = \{z:|z|=1\}$$ Suppose z is a root of 1, i.e, $$z=e^{2\pi i p/q}$$ , $$p, q \in \mathbb Z, q>0$$ For $$n\geq q$$ q | n! and $$z^{n!}=1$$ Next is what I don't get. (continue on comment)

1. Loser66

So, $$\sum_{n =1}^\infty \dfrac{(-1)^n}{n}z^{n!}=\sum_{n=1}^{q-1} \dfrac{(-1)^n}{n}z^n +\sum_{n =q}^{\infty} \dfrac{(-1)^n}{n}$$

2. Loser66

The first sum is finite and so does not affect convergence The second one is an alternating harmonic series which converges. Then the whole thing converges. I don't get it. Please, help how can the sum be break into 2 parts like that?

3. Loser66

@ganeshie8

4. Loser66

nvm, I got it.