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Find all z such that the series zn / n1/2 * log ((2n+1) / n) converges

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \frac{z^n}{\sqrt{n}} \log \frac{2n+1}{n} \]

converges.


First, we simplify the expression as follows,

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

Then, we consider the first series,

    \[ \log 2 \cdot \sum_{n=1}^{\infty}\frac{z^n}{\sqrt{n}}. \]

Let a_n = \frac{z^n}{\sqrt{n}} and use the ratio test,

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

This series converges if |z| < 1 and diverges if |z| > 1. If |z|=1 and z \neq 1 then the series also converges by Dirichlet’s test. Finally, if z = 1, then the series is \sum \frac{1}{\sqrt{n}} which diverges.

Next, we know

    \[ \log \left( 1 + \frac{1}{2n} \right) < \log 2 < 1 \qquad \text{for all } n \geq 1. \]

Thus,

    \begin{align*}  \sum_{n=1}^{\infty} \log \left( 1 + \frac{1}{2n} \right) \leq \sum_{n=1}^{\infty} \frac{z^n}{\sqrt{n}}. \end{align*}

Hence, this series converges anywhere \sum_{n=1}^{\infty} \frac{z^n}{\sqrt{n}} converges. Therefore, the sum

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

converges for |z| \leq 1 with z \neq 1.

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