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Find all z such that the series (1 + 1/(5n+1))n2 |z|17n converges

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \left( 1 + \frac{1}{5n+1} \right)^{n^2} |z|^{17n} \]

converges.


We look to apply the root test, letting

    \[ a_n = \left( 1 + \frac{1}{5n+1} \right)^{n^2} |z|^{17n}. \]

Then we have

    \begin{align*}  \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( 1 + \frac{1}{5n+1} \right)^n |z|^{17} \\[9pt]  &= \lim_{n \to \infty} e^{\frac{1}{5}} \cdot |z|^{17}. \end{align*}

Since this series converges if this limit is less than 1 and diverges if it is greater than 1, we have the series is absolutely convergent for

    \[ e^{\frac{1}{5}} \cdot |z|^{17} < 1 \quad \implies \quad |z|<e^{-\frac{1}{85}}. \]

3 comments

  1. Rafael Deiga says:

    And about |z|=e^(-1/85)? In this case, I know that the root test is inconclusive, but is there a way to guarantee the convergence or divergence?

    • Rafael Deiga says:

      With |z|=e^(-1/85), the limit of the terms of the serie goes to e^(-3/50) as n-> infinity. Therefore, the series diverge in this case.

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