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

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

by RoRi

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.

4 comments

  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…

    • Artem says:

      I do not think these answers are correct, it should be solved somehow else. The complex sequence cannot indeed be decreasing

    • Anonymous says:

      Hello!

      It is indeed odd to state that such a complex sequence is monotonic. If you are looking for help with the solution. Take the series of absolute values and show that it converges (for example with the ratio test) for |z-1| 1 show that the limit a_n \to 0 as n \to \infty is not true, where a_n is the n-th term of the sum.

      Cheers!

      • Anonymous says:

        The previous comment got concatenated so here is the rest.

        For |z-1| = 1 use Drichlet’s test to conclude that it converges for the case when the arg|z-1| \ne 2\pi k, k\in Z and Leibnitz’s test when arg|z-1| = 2\pi k, k\in Z. The |z-1|1 in the previous comment is |z-1| < 1.

        Cheers!

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):