Question

Decide whether the series is convergent or divergent

A)) ∑ [ 8^{ n+i.(2^-n) } ] / [ 9 ^ n ]

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B)) ∑ Conjugate all of this ( [ n+in+(n^i) ] / [ i ^ n ] )

Answer 1

Answer on Question #50227 – Math – Complex Analysis

Decide whether the series is convergent or divergent

A)) $\sum [ 8^{\wedge} \{ n + i.(2^{\wedge} - n) \} ] / [ 9^{\wedge} n ]$

B)) $\sum$ Conjugate all of this ($[ n + \mathrm{in} + (n^{\wedge}i) ] / [ i^{\wedge} n ]$)

Solution

A) $\sum \frac{8^{n + i2^{-n}}}{9^n} = \binom{8}{9}^n 8^{i2^{-n}} \sum \frac{8^{n + i2^{-n}}}{9^n} = \sum \binom{8}{9}^n 8^{i/2^n}.$

By Cauchy criterion

$$q = \lim_{n \to \infty} \sqrt[n]{\left| \binom{8}{9}^n 8^{i/2^n} \right|} = \frac{8}{9} < 1, \text{ hence the series is convergent.}$$

B) $\sum \left(\frac{n + \mathrm{in} + n^i}{i^n}\right) = \sum \left(\frac{n + \mathrm{in} + e^{ilnn}}{i^n}\right)$. By Cauchy criterion

$$\begin{aligned} q = \lim_{n \to \infty} \sqrt[n]{\left| \frac{n + \mathrm{in} + e^{ilnn}}{i^n} \right|} &= \lim_{n \to \infty} |n + \mathrm{in} + e^{ilnn}| \\ &= \lim_{n \to \infty} \sqrt{(n + \cos(lnn))^2 + (n + \sin(lnn))^2} = +\infty, \text{ hence} \end{aligned}$$

the series is divergent.

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