4.1 參考解答

3. $\lim_{n \to \infty} z_n=z_0$, $\lim_{n \to \infty} \overline{z_n}=\overline{z_0}$

   By $\forall \space \varepsilon$, $\exist N_\varepsilon>0$ such that $n>N_\varepsilon$, $|z_n-z_0|<\varepsilon$

   $\therefore \forall \space \varepsilon$, $\exist \space N=N_\varepsilon$ such that $\forall \space n>N$, $|\overline{z_n}-\overline{z_0}|=|\overline{z_n-z_0}|=|z_n-z_0|<\varepsilon$

   $\therefore \lim_{n \to \infty} \overline{z_n}=\overline{z_0}$

5. $\Sigma_{n=0}^{\infty} (\frac{1}{n+1+i}-\frac{1}{n+i})=i$

10. (a) $\Sigma_{n=1}^{\infty} \frac{(i)^n}{n}=i+\frac{-1}{2}+\frac{-i}{3}+\frac{1}{4}+\frac{i}{5}+\frac{-1}{6}+\frac{-i}{7}+\frac{1}{8}+\cdots$

    $=(-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\frac{1}{8}+-\cdots)+i(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+-\cdots)$

    $=\Sigma_{n=1}^{\infty} \frac{(-1)^n}{2n}+i\Sigma_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$

    $\Sigma_{n=1}^{\infty} \frac{(-1)^n}{2n}$: $1^。$ by alternatintest $2^。$ $\lim_{n \to \infty} \frac{1}{2n}=0$ : converges

    同理 $\Sigma_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$ is also convergent.

    $\therefore \Sigma_{n=1}^{\infty} \frac{(i)^n}{n}$ converges.

    (b) $\Sigma_{n=1}^{\infty} (\frac{1}{n}+\frac{i}{2^n}) \longrightarrow$ diverges

    Let $x_n=\frac{1}{n}$, $y_n=\frac{i}{2^n}$

    $\Sigma_{n=1}^{\infty} x_n=\Sigma_{n=1}^{\infty} \frac{1}{n}$ is divergent. (by $p$-series)

12. $|\Sigma_{n=0}^{k} z_n|=|z_0+z_1+z_2+\cdots +z_k|$