Quiz 2 Session 6101

[Jiaqi Bi]

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.

[Pengyun Li]

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.