Toronto Math Forum
MAT334--2020F => MAT334--Tests and Quizzes => Quiz 2 => Topic started by: Jiaqi Bi on October 01, 2020, 04:46:51 PM
-
Question: $$\sum_{n=1}^{\infty} \frac{1}{n} (\frac{1+i}{\sqrt{2}})^{n}$$
Answer:
$$
\lim_{n \to \infty} \frac{1}{n}(\frac{1+i}{\sqrt{2}})^{n} \\
=\lim_{n \to \infty} \frac{1}{n}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\
=\lim_{n \to \infty} \frac{1}{n}\cdot \lim_{n \to \infty}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\
$$
Since $$\lim_{n \to \infty}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\
=\lim_{n \to \infty}(\frac{\sqrt{2}\ (cos\frac{\pi}{4}+isin\frac{\pi}{4})}{\sqrt{2}})^{n}\\
=\lim_{n \to \infty}(cos \frac{n\pi}{4}+isin \frac{n\pi}{4}) \ \ (\text{By De Moivres Theorem})\\
=\text{DNE}\ \text{and it diverges}
$$
Hence, the series diverges.
-
I think the answer should be convergent instead? But correct me if I'm wrong :)
$(\frac{1+i}{\sqrt{2}})^n = e^{i\frac{n\pi}{4}} = \cos (\frac{n\pi}{4})+i\sin (\frac{n\pi}{4})$.
When $n = 1$, $\frac{1+i}{\sqrt{2}} = \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$.
When $n = 2$, $(\frac{1+i}{\sqrt{2}})^2 = 0+i$.
When $n = 3$, $(\frac{1+i}{\sqrt{2}})^3 = -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$.
......
Then you will find that the result cycles for every 8 members.
Denote the result for $n=1$ as $Z_1$, $n=2$ as $Z_2$ , etc.,
$Z_1 + Z_5 + Z_9 +.... = (\frac{1}{\sqrt{2}}-\frac{1}{5\sqrt{2}}+\frac{1}{9\sqrt{2}}-....)+i(\frac{1}{\sqrt{2}}-\frac{1}{5\sqrt{2}}+\frac{1}{9\sqrt{2}}-.....)$. It is very obvious to see that this series is convergent.
Similarly, we can figure out that $(Z_2 + Z_6 + Z_{10} +.... )$, $(Z_3 + Z_7 + Z_{11}+.... )$, $(Z_4 + Z_8 + Z_{12} +.... )$ all convergent.
Thus, $\sum_{n=1}^{\infty}(\frac{1+i}{\sqrt{2}})^n$ is convergent.
Also, $\lim_{n\to\infty} \frac{1}{n}\rightarrow 0$, is convergent.
Therefore, $\sum_{n=1}^\infty \frac{1}{n}(\frac{1+i}{\sqrt{2}})^n$ is convergent.