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I tried to show that $s$ was absolute convergent but it was divergent so that gave nothing.

Next i tried to write the sum $s$ as four different sums by using the fact that $i^n$ rotates through four different values (i,-1,-i,1):

$s = s_1+s_2+s_3+s_4 = \sum_{n=1}^{\infty} \frac{i^n}{n} $ = $\sum_{k=1}^{\infty} \frac{1}{4k} $ + $\sum_{k=1}^{\infty} \frac{-1}{4k-2} $ + $\sum_{k=0}^{\infty} \frac{i}{4k+1} $ + $\sum_{k=1}^{\infty} \frac{-i}{4k-1} $

From here, in order to show that $s$ is convergent I need to show that the four sums are together convergent. I see directly that $s_1$ and $s_2$ are divergent (p-series with p=1) so for $s$ to be convergent $s_3$ and/or $s_4$ must be divergent to "cancel out" the divergence of $s_1$ and $s_2$. I don't know how to show that they cancel out each other so I am stuck here.

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  • $\begingroup$ I guess you can treat $i$ as a constant since $i^1=I,i^2=-1,i^3=-i,i^4=1$ and then it repeats the same if you multiply a scalar for the powers. So you take the $\lim_{\infty} \frac{i^n}{n}=0$. $\endgroup$ – Pedro Gomes Dec 1 '17 at 18:51
  • $\begingroup$ Hint: $\sum_{n=1}^\infty (-1)^n / (2n)$ converges. $\endgroup$ – Reiner Martin Dec 1 '17 at 18:53
  • $\begingroup$ You can't split it the way you wrote because it's conditionally convergent, and doing what you did gets you to summation of 4 divergent series. That's because the sum of absolute values diverges. In these cases you must not mess with the order of terms too much. $\endgroup$ – orion Dec 1 '17 at 19:22
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$s = \sum_\limits{n=1}^\infty \frac {(-1)^{n}}{2n} + i\sum_\limits{n=0}^\infty \frac {(-1)^{n}}{2n+1}$

And each of those series are conditionally convergent

$s = -\frac 12 \ln 2 + i\frac \pi 4$

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Just apply Dirichlets' test and use the fact that $\sum_{k=1}^Ni^k\in\{i,-1+i,-1,0\}$, for each $N\in\mathbb N$.

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  • $\begingroup$ We are not supposed to use this test as it hasn't been introduced yet in the course. We are supposed to use the calculus theorems (comparison test, divergence test, integral test etc..) and/or the fact that both the imaginary and real part of the sum must be convergent for the sum to be convergent. $\endgroup$ – SwedeGustaf Dec 1 '17 at 18:57
  • $\begingroup$ @Gustaf Were we supposed to guess that from your question? $\endgroup$ – José Carlos Santos Dec 1 '17 at 18:58
  • $\begingroup$ No sorry, my bad! $\endgroup$ – SwedeGustaf Dec 1 '17 at 19:01
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The even terms of your sum form an alternating series. The odd terms are $i$ times an alternating series. The terms in both series decrease to $0$, so the sum converges.

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  • $\begingroup$ Good answer. That helped! $\endgroup$ – SwedeGustaf Dec 1 '17 at 18:58
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We can write, $\sum_{n=1}^{\infty}(\iota^n/n)= \sum_{k=1}^{\infty}((-1)^k/2k) + \iota \sum_{k=1}^{\infty}((-1)^{k+1}/(2k-1))$ which are convergent by Leibniz Test.

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