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Find all z such that the series nnzn converges

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} n^n z^n \]


We consider the limit of the nth root of the terms of the series,

    \begin{align*}  \lim_{n \to \infty} (a_n)^{\frac{1}{n}} &= \lim_{n \to \infty} \left( n^n z^n \right)^{\frac{1}{n}} \\[9pt]  &= \lim_{n \to \infty} nz \\[9pt]  &= \begin{dcases} +\infty &\text{if } z \neq 0 \\ 0 & \text{if } z = 0. \end{dcases} \end{align*}

Hence, the series converges if and only if z = 0.


  1. Artem says:

    Is it the root test used here? I do not think that is legal, since the root test is defined only for real numbers in Apostol (it requires a sequence of non-negative terms, which nz clearly does not follow).

    • Anonymous says:


      Yes, I think it’s a mistake too, as an alternative it can be proved that except when z=0 the necessary condition of convergence \lim\limits_{n \to \infty} a_n = 0 is not satisfied.
      When z=0, if we want to be pedantic, we can show that the limit of partial sums is 0 as n goes to \infty.


    • S says:

      But if you consider absolute convergence, it is positive and it is real. Because absolute convergence implies the conditional one, the proof seems valid to me.

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