Question

Decide whether this sequence is convergent or divergent:

Bn = [ Ln (2-n) ] / [ | n +i {square root (2) | ^ {2} ] ,
n≥ 3

Note : please Ln is principle value
Please more than method if could appreciated

Answer 1

Answer on Question #50126, Math, Complex Analysis

Given:

$$B_n = \frac{Ln(2 - n)}{(|n + i\sqrt{2}|)^2} \qquad n \geq 3$$

Decide:

whether these series is convergent or divergent

Solution:

$$B_n = \frac{Ln(2 - n)}{(|n + i\sqrt{2}|)^2} = \frac{Ln|2 - n| + i(\arg(2 - n) + 2\pi k)}{(\sqrt{n^2 + 2})^2}$$ $$n \geq 3 \quad \Rightarrow \quad 2 - n \leq 0 \quad \Rightarrow \quad \arg(2 - n) = \pi$$

$Ln$ is a principle value, so $k = 0$

$$B_n = \frac{Ln(n - 2) + i\pi}{n^2 + 2} \quad |B_n| = \frac{\sqrt{Ln^2(n - 2) + \pi^2}}{n^2 + 2} \sim \frac{Ln(n - 2)}{n^2}$$

Using the integral test for convergence by Maclaurin Cauchy

$$\begin{aligned} \int_{3}^{+\infty} \frac{Ln(x - 2)}{x^2} dx &= \left\{u = Ln(x - 2), v' = \frac{1}{x^2}, u' = \frac{1}{x - 2}, v = -\frac{1}{x}\right\} = \\ &= -\frac{Ln(x - 2)}{x} \Bigg|_{3}^{+\infty} + \int_{3}^{+\infty} \frac{1}{x(x - 2)} dx = \left(\frac{Ln(x - 2)}{x} + \frac{1}{2}[Ln(x - 2) - Ln(x)]\right)\Bigg|_{3}^{+\infty} = \\ &= \left(- \frac{Ln(x - 2)}{x} + \frac{1}{2}Ln \frac{x - 2}{x}\right)\Bigg|_{3}^{+\infty} = \frac{Ln(3 - 2)}{3} - \frac{1}{2} \frac{Ln(3 - 2)}{3} = -\frac{1}{2}Ln \frac{1}{3} \quad \in R \end{aligned}$$

Answer: convergent

www.assignmentexpert.com