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Find all z such that the series (-1)n(z-1)n / n converges

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^n (z-1)^n}{n} \]

converges.


If |z-1| < 1 then we apply the Leibniz rule, where \frac{(z-1)^n}{n} is monotonically decreasing and has limit 0 as n \to \infty. If |z-1| = 1 with z \neq 0, then we apply Dirichlet’s test with b_n = \frac{1}{n} to conclude that the series converges. If z = 0 then (z-1)^n = (-1)^n and the series diverges since it equals the harmonic series.

One comment

  1. tom says:

    Regarding Leibniz’ rule, I don’t understand how a complex sequence is defined as monotonic. I can see how a power series has a radius of convergence < 1 though. Perhaps I'm just tired…

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