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Consider the series, $$\sum_{n=2}^{\infty}\frac{i^n}{\ln{(n)}}.$$ I would like to test this series for convergence. I tried doing the ratio test. We have that, $$\bigg|\frac{c_{n+1}}{c_n}\bigg|=\bigg|-i\frac{\ln{(n+1)}}{\ln{(n)}}\bigg|=\frac{\ln{(n+1)}}{\ln{(n)}}.$$ However, I'm not so sure this helps me in finding out whether or not the series is convergent. I can't tell if the ratio of the logarithms is less than or greater than $1$. Are there any other ways to approach the problem (i.e. any other tests for convergence I could do) in this case?

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    $\begingroup$ In this case the ratio test is inconclusive. It does not tell you anything, since the limit is $1$. $\endgroup$ – Crostul Mar 17 '18 at 13:33
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You can split your series in its real and imaginary parts in the following way:

$\sum_{n=2}^{\infty}\frac{i^n}{\ln{(n)}}=\sum_{n=1}^{\infty}\frac{(-1)^n}{\ln{(2n)}}+i\sum_{n=1}^{\infty}\frac{(-1)^n}{\ln{(2n+1)}}$

Then you can use the alternating series test on the two series. Hope this helps.

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HINT: observe that

$$\sum_{k=2}^\infty\frac{i^k}{\ln k}=\lim_{n\to\infty}\sum_{k=2}^n\frac{i^k}{\ln k}=\lim_{n\to\infty}\sum_{k=2}^n\frac{\Re(i^k)+\Im(i^k)}{\ln k}=\lim_{n\to\infty}\left(\sum_{k=2}^n\frac{\Re(i^k)}{\ln k}+\sum_{k=2}^n\frac{\Im(i^k)}{\ln k}\right)$$

Then if you shows that the real part and the imaginary part of the series are conditionally convergent you are done, that is, if $\lim a_k=L_1$ and $\lim b_k=L_2$ then $\lim(a_k+b_k)=L_1+L_2$.

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