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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.

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

Find all complex numbers z such that the series

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

converges.


Let

    \[ b_n = \frac{1}{2n-1}, \qquad a_n = \left( \frac{1-z}{1+z} \right)^n. \]

Then the series converges if \sum a_n converges. But, since \sum a_n is a geometric series, it converges if

    \[ \left| \frac{1-z}{1+z} \right| < 1 \quad \implies \quad |1-z| < |1+z|. \]

If we write z = x + iy this inequality holds if and only if x > 0.
On the other hand,

    \[ |1-z| > |1+z| \quad \implies \quad \frac{1-z}{1+z} > 1 \]

and so

    \[ \lim_{n \to \infty} \frac{1}{2n-1} \left( \frac{1-z}{1+z} \right)^n \neq 0 \]

hence, the series diverges.

Therefore, the series converges for all z = x + iy with x \geq 0.

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.

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

Find all complex numbers z such that the series

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

converges.


First, the series is not absolutely convergent for any z \in \mathbb{Z} since

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

By limit comparison with \sum \frac{1}{n} this always diverges.

Then, if z \neq -n for any positive integer n the series is convergent since \frac{1}{z+n} is monotonically decreasing and \lim_{n \to \infty} \frac{1}{z+n} = 0. If z = -n for some positive integer n, then the series is not defined since it is undefined for that term.