回復 30# beaglewu 的帖子
第2題
\(\begin{align}
& {{S}_{n}}=\sum\limits_{k=1}^{n}{\left| {{Z}_{k+1}}-{{Z}_{k}} \right|} \\
& =\sum\limits_{k=1}^{n}{{{\left| \frac{1-i}{2} \right|}^{k}}\left| \frac{1-i}{2}-1 \right|} \\
& ={{\sum\limits_{k=1}^{n}{\left( \frac{\sqrt{2}}{2} \right)}}^{k+1}} \\
& \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,{{\sum\limits_{k=1}^{n}{\left( \frac{\sqrt{2}}{2} \right)}}^{k+1}}=\frac{\frac{1}{2}}{1-\frac{\sqrt{2}}{2}}=\frac{2+\sqrt{2}}{2} \\
\end{align}\)
