Question

Sequence Zn = [ e ^{ i.(n ^2)- (n^2)} ] ^ (1/n)
A )) Determine whether Zn is convergent or divergent
Note : please by find the Limit Zn :as n approach to infinity: ( if value : the sequence will convergent ,but if Does not exist or infinity the sequence is divergent

B )) is ∑ Zn ^ n Convergent ? Why?

Answer 1

Answer on Question #50268 – Math – Complex Analysis

Sequence Zn = [ e ^ { i.(n ^ {2}) - (n ^ {2}) } ] ^ { (1 / n) } 


A )) Determine whether Zn is convergent or divergent

Note : please by find the Limit Zn :as n approach to infinity: ( if value : the sequence will convergent ,but if Does not exist or infinity the sequence is divergent

B )) is $\sum \mathrm{Zn}^{\wedge}\mathrm{n}$ Convergent ? Why

Solution

A. $Z_{n} = \left(\exp \left[in^{2} - n^{2}\right]\right)^{1 / n}$

$\lim_{n\to \infty}\left|Z_n\right| = \lim_{n\to \infty}\left|\left(\exp \left[in^{2} - n^{2}\right]\right)^{1 / n}\right| = \lim_{n\to \infty}\left|\left(\exp \left[-n^{2}\right]\right)^{1 / n}\right| = \lim_{n\to \infty}\left|e^{-n}\right| = 0 < 1$ , hence the series is convergent.

B. $Z_{n}^{\prime} = \left(Z_{n}\right)^{n} = \left(\exp \left[in^{2} - n^{2}\right]\right)$

$\lim_{n\to \infty}\left|Z_n'\right| = \lim_{n\to \infty}\left|\left(\exp \left[in^{2} - n^{2}\right]\right)\right| = \lim_{n\to \infty}\left|\left(\exp \left[-n^{2}\right]\right)\right| = \lim_{n\to \infty}\left|e^{-n^{2}}\right| = 0 < 1$ , hence the series is convergent.

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