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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)
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)

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0So, \(\sum_{n =1}^\infty \dfrac{(1)^n}{n}z^{n!}=\sum_{n=1}^{q1} \dfrac{(1)^n}{n}z^n +\sum_{n =q}^{\infty} \dfrac{(1)^n}{n}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The 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?
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