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Find all z such that the series (n / (n+1)) (z / (2z+1))n converges

Find all complex numbers z such that the series

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

converges.


First, we have

    \begin{align*}  \sum_{n=1}^{\infty} \left( \frac{n}{n+1} \right) \left( \frac{z}{2z+1}\right)^n &= \sum_{n=1}^{\infty} \left( 1 - \frac{1}{n+1} \right)\left( \frac{z}{2z+1} \right)^n \\[9pt]  &= \sum_{n=1}^{\infty} \left( \left( \frac{z}{2z+1} \right)^n - \left( \frac{1}{n+1}\right)\left( \frac{z}{2z+1}\right)^n \right). \end{align*}

We know from the previous exercise (Section 10.20, Exercise #44) that the series

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

converges if and only if

    \[ \left| 2+ \frac{1}{|z|} \right| > 1. \]

Therefore, the series converges if \left| 2 + \frac{1}{|z|} \right| > 1 since both terms in the sum converge. The series diverges if \left| 2 + \frac{1}{|z|} \right| = 1 since \sum \left( \frac{z}{2z+1} \right)^n diverges and \sum \frac{1}{n} \left( \frac{z}{2z+1} \right)^n converges. Finally, the series diverges if \left| 2 + \frac{1}{|z|} \right| = < 1 since \lim_{n \to \infty} a_n \neq 0.

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