determine whether the series convergence
$\sum _{n=1}^{\infty \:}\frac{i^n}{n} $
My teacher said it is convergent but the ratio test is inconclusive and the root test is inconclusive
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$\begingroup$ Does the sequence of partial sums converge? You may have to look at a few -- say, 4 -- subsequences. If they all converge, and to the same limit, then... $\endgroup$– Clement C.Nov 7 '16 at 21:31
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1$\begingroup$ @ClementC. $4$? Isn't $2$ better? $\endgroup$– b00n heTNov 7 '16 at 21:32
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$\begingroup$ Whatever works, I'd have gone for 4 (out of safety). $\endgroup$– Clement C.Nov 7 '16 at 21:33
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The series is convergent but not absolutely convergent. Absolute convergence would say that the series $$ \sum_1^\infty \left| \frac{i^n}{n} \right| = \sum_1^\infty \frac{1}{n} $$ converges, and we know that is not the case.
However, we can break the series in question up as $$ \sum_{n=1}^\infty \frac{i^n}{n} = \sum_{m=1}^\infty (-1)^m \frac{1}{2m} + i \sum_{m=0}^\infty (-1)^m \frac{1}{2m+1} $$ and each of those alternating sign series can be shown to converge by grouping two terms together, getting a sum like $$ \sum_{m=1}^\infty (-1)^m \frac{1}{2m} = \frac12 \sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+1} \right) = \frac12 \sum_{k=1}^\infty \frac1{k^2+k} $$ which converges by the ratio test.
answered Nov 7 '16 at 21:39
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$\begingroup$ The last lines are unnecessary if you quote Leibniz Series Theorem. +1 $\endgroup$ Nov 7 '16 at 21:55
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Hint: Subdivide the Series in two sub-series: one over $2n$ and one over $2n+1$ and show that these converge.
