I would like to show that the series $$ \sum_{n=1}^\infty \frac{i^n}{n} $$ converges.
My first idea was to use ratio test but that does not help since $$ \left|\frac{a_{n+1}}{a_n}\right|=\frac{n}{n+1}\to 1 $$
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2A test that uses absolute values will not work on this, as the series isn't absolutely convergent. – Arthur Nov 30 '22 at 22:05
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See https://math.stackexchange.com/q/996553/42969 – Martin R Nov 30 '22 at 22:05
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1Break the series as $\sum (-1)^ka_k+i\sum (-1)^kb_k$ where $a_k,b_k$ are positive reals decreasing to $0,$ so they converge by the alternating series rule. – Thomas Andrews Nov 30 '22 at 22:10
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Does this answer your question? Determining precisely where $\sum_{n=1}^\infty\frac{z^n}{n}$ converges? – Anne Bauval Nov 30 '22 at 23:25
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4(Yes, it did.) This also answers your question: Is $\sum{\frac{i^{n}}{n}}$ convergent or divergent?. – Anne Bauval Jan 09 '23 at 12:53
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Consider the below series: $$\sum _{n=1}^{\infty }\frac{x^n}{n}=-\ln \left(1-x\right)$$
$$x=i\Rightarrow \sum _{n=1}^{\infty }\frac{i^n}{n}=-\ln \left(1-i\right)=-\frac{1}{2}\ln \left(2\right)+\frac{i\pi }{4}$$ Finally: $$\sum _{n=1}^{\infty }\frac{i^n}{n}=-\frac{1}{2}\ln \left(2\right)+\frac{\pi }{4}i$$ note: $$\ln \left(a+bi\right)=\frac{1}{2}\ln \left(a^2+b^2\right)+i\arctan \left(\frac{b}{a}\right)$$
asmatqatea
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This answer is correct but "cheating": the first line suffices to answer much more than what was asked, but the knowledge it implicitly assumes does not correspond to the question's level (I wonder if the accepter fully understood it). Anyway, that question was a (multi-)duplicate. – Anne Bauval Jan 09 '23 at 13:02