### Author Topic: Test 1 Question 2  (Read 675 times)

#### Jiayu Chen

• Newbie
• Posts: 1
• Karma: 0
##### Test 1 Question 2
« on: November 16, 2020, 11:20:31 AM »
Can somebody help me with question2? I lost mark in calculating R.

#### RunboZhang

• Sr. Member
• Posts: 51
• Karma: 0
##### Re: Test 1 Question 2
« Reply #1 on: November 16, 2020, 01:47:37 PM »
$\textbf{For part (a):} \\\\ \text{By ratio test, we have : }\\\\$

\begin{gather} \begin{aligned} \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty}|\frac{(1+ i\sqrt{3})^{n+1}}{4^{n+1}(n+1)log^{3}(n+1)} \cdot \frac{n 4^{n} log^{3}(n)}{(1+i\sqrt{3})^{n}}| \\\\ &= \lim_{n \to \infty} |\frac{1+i\sqrt{3}}{4} \cdot \frac{nlog^{3}(n)}{(n+1)log^3(n+1)}| \\\\ &= \frac{1}{2} \text{(by further expanding this absolute value)}\\\\ &= \frac{1}{R} \end{aligned} \end{gather}

$\text{Therefore the radius of convergence is 2.}$

$\textbf{For part (b):} \\\\ \text{We can follow the similar method. By ratio test, we have : }\\\\$

\begin{gather} \begin{aligned} \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty} |\frac{(i+\sqrt{3})^{n+1} \cdot \frac{e^{2(n)+1}-1}{e^{2(n+1)}+1}}{4^{n+1}} \cdot \frac{4^n}{(i+\sqrt{3})^{n}\cdot \frac{e^{2n}-1}{e^{2n}+1}}| \\\\ &= \lim_{n \to \infty} |\frac{i+\sqrt{3}}{4} \cdot \frac{e^{2n+2}-1}{e^{2n+2}+1} \cdot \frac{e^{2n}+1}{e^{2n}-1}| \\\\ &= \frac{1}{2} \text{(by further expanding this absolute value, and most terms can be canceled out.)}\\\\ &= \frac{1}{R} \end{aligned} \end{gather}

$\text{Therefore the radius of convergence is 2.} \\\\$

$\text{Above is how I got my answers. Correct me if I made any mistakes. Feel free to comment below }$
« Last Edit: November 16, 2020, 01:53:28 PM by RunboZhang »