Question

B Decide whether these series is convergent or divergent .


1. ∑ (n+i sin n ) ^ {1/n}
2 . ∑ ( n^{in} ) , where n ^ in is taken in the 3π -branch

Hint : If you wish,You can use the fact
lim(n→∞) Zn=0 <--> lim(n→∞) |Zn| = 0

Answer 1

Answer on Question #50130 - Math - Complex Analysis

Absolute value of $a_n = n + i \sin n$ is greater than 1, so the absolute value of $(n + i \sin n)^{1/n}$ is also greater than 1 and hence $a_n$ does not converge to 0. So the initial series isn't convergent.

$n^{in} = (e^{\ln n + 2\pi i})^{in} = e^{in \ln n} e^{-2\pi n}$ that its absolute value converges to zero exponentially. Therefore, by comparison criterion, the initial series converges.

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