Is the next series absolutely convergent, semi convergent or divergent? $$\sum_{n\ge1}{\frac{i^n}{n}}$$ where $i$ is the complex number such that $i^2=-1$. I don't know that I can use Leibniz Criterion because there is no $(-1)^n$, it's a complex noumber instead. Someone told me that I should use Dirichlet's test but I don't realy know how. Can somebody put me on the right track?
Chck if the series $\sum_{n\ge1}{\frac{i^n}{n}}$ absolutely convergent, semi convergent or divergent?
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$\begingroup$
calculus
convergence
absolute-convergence
2 Answers
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Hint:
- A series $$\sum_{n=1}^\infty (a_n + b_n i)$$ where $a_n, b_n\in\mathbb R$ is convergent if and only if both the series $$\sum_{n=1}^\infty a_n, \sum_{n=1}^\infty b_n$$ are convergent.
- In your case, $|a_n| = \frac1n$ if $n$ is even and $0$ if it is odd, while $|b_n|=\frac1n$ if $n$ is odd and $0$ if it is even. Also, look at the signs of $a_n$ and $b_n$.
- The series is absolutely convergent if and only if the series $$\sum_{n=1}^\infty\left|\frac{i^n}{n}\right|$$ is convergent.
- $|i^n| = |i|^n = 1$
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Hint: Absoute convergence is trivially wrong. To prove convergence separate odd and even terms.
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0I did not said that is Absolut convergent, I have to check that.. And If it is not absolute convergent I shouldn't be alowed to change the position of the terms inside of it, right? So I can't separete the odd and even terms? – 2017-01-09
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0what I meant is: it is trivially not absolutely convergent. For the other "problem" see 5xum's answer. – 2017-01-09