Question

Find if these sequence convergent of divergent ( details )

1) [ (1+i)^(1/n) ] / square root(n-1)
2) [2^n! ] / [2^ (n+1)! ]
3) sin ((1+i)/n)

Answer 1

Answer on Question #49484 – Math – Complex Analysis

Find if these sequence convergent of divergent (details)

1) $[(1 + \mathrm{i})^{\wedge}(1 / \mathrm{n})] / \text{square root}(\mathrm{n - 1})$

2) $[2^{\wedge}\mathrm{n}!] / [2^{\wedge}(\mathrm{n} + 1)!]$

3) $\sin ((1 + \mathrm{i}) / \mathrm{n})$

Solution

We say that $(z_{n})$ converges to $A$ (write $z_{n}\to A$ or $\lim_{n\to \infty}z_n = A$), if for every real number $\varepsilon$, there exists a natural number $N$ such that

$$n \geq N \Rightarrow | z _ {n} - A | < \varepsilon$$

1) $\left|\frac{(1 + i)^{\frac{1}{n}}}{\sqrt{n - 1}}\right| = \frac{\frac{1}{2^{2n}}}{\sqrt{n - 1}} \leq \frac{\sqrt{2}}{\sqrt{n - 1}} < \frac{2}{\sqrt{n - 1}} \to 0$ as $n \to \infty$, besides, $2^{\frac{1}{2n}} \to 2^0 = 1$ as $n \to \infty$, $\sqrt{n - 1} \to \infty$ as $n \to \infty$. We take for any $\varepsilon > 0$ natural number $N = \left[\left(\frac{2}{\varepsilon}\right)^2\right] + 1$ such that for all $n \geq N \Rightarrow \frac{2}{\sqrt{n - 1}} < \varepsilon$, hence

$\left|\frac{(1 + i)^{\frac{1}{n}}}{\sqrt{n - 1}}\right| < \varepsilon$. Thus, the sequence $z_{n} = \frac{(1 + i)^{\frac{1}{n}}}{\sqrt{n - 1}}$ is convergent.

2) $\frac{2^{n!}}{2^{(n + 1)!}} = 2^{(n! - (n + 1)!)} = 2^{n!(1 - (n + 1))} = 2^{-n\cdot n!} = \frac{1}{2^{n\cdot n!}} \to 0$ as $n \to \infty$, so the sequence $z_{n} = \frac{2^{n!}}{2^{(n + 1)!}}$ with positive terms is convergent.

3) $\left|\sin \left(\frac{1 + i}{n}\right)\right| = \left|\frac{e^{i\left(\frac{1 + i}{n}\right)} - e^{-i\left(\frac{1 + i}{n}\right)}}{2i}\right| \to 0$ as $n \to \infty$, because $\frac{1 + i}{n} \to 0$ as $n \to \infty$ (here $\left|\frac{1 + i}{n}\right| < \frac{2}{n} \to 0$ as $n \to \infty$). Thus, the sequence $z_{n} = \sin \left(\frac{1 + i}{n}\right)$ is convergent.

www.AssignmentExpert.com