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Study the convergence of $$\sum_{n \geq 2 } \frac{i^n}{\log(n)} $$ where $\log$ of course denotes the 'natural logarithm' and $i \in \mathbb{C}$

Oddly enough I managed to show Abel's Test for convergence (using Cauchy Criteria) but when it comes to applying it to the above sum I get stuck. $$(a_n):= \frac{1}{\log(n)} $$ is indeed a monotone decreasing sequence in $\mathbb{R}$ that converges to $0$, thus I define $$(B_n):= \sum_{k=1}^n i^k $$ and I want to show that this sequence $(B_n)_{ n \geq 1}$ is bound independently of $n$, computing a few values of the sequences I get: $$B_n = \lbrace i, i-1,-1,0,i \rbrace \text{ for } n=1,2,3,4,5 $$

And for bigger values of $n$ I don't seem to get additional values for the list, the sum repeats, so I should feel comfortable.


My Problem: In the proof of Abel's Test I made use of the fact that $(B_n)$ is bounded independently on $n$. So Although we're dealing with complex numbers I said there exists $B \in \mathbb{C}$ such that $|B_n| \leq B$ for all $n \in \mathbb{N}$.

So I of course I want to show that the same applies to the $(B_n)$ as defined above. But simply saying that the sum is either $i,i-1,-1,0,i$ doesn't feel satisfying to me as well. I would like there to be a statement that I could maybe show via induction, am I really supposed to show $$A(n): \sum_{k=1}^n i^k = i \text{ or } i-1 \text{ or }... \forall n \in \mathbb{N} $$

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  • $\begingroup$ Recall the formula for geometric sums, if $q\neq 1$, then $$\sum_{k=0}^n q^k = \,?$$ $\endgroup$ – Daniel Fischer Oct 20 '14 at 21:22
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    $\begingroup$ Also, you could separate the real and imaginary parts (even and odd indices), and apply the Leibniz criterion to each. $\endgroup$ – Daniel Fischer Oct 20 '14 at 21:24
  • $\begingroup$ I had the part of the geometric series already written down on paper @DanielFischer but then tossed it away because the sum would depend on $n$ wouldn't it? And in the theorem we have to find a bound independent of $n$ or do I misunderstand things here? $\endgroup$ – Spaced Oct 20 '14 at 21:26
  • $\begingroup$ The sum depends on $n$, but you can find a bound independent of $n$ using the sum formula. $\endgroup$ – Daniel Fischer Oct 20 '14 at 21:27
  • $\begingroup$ I believe my problem is the argumentation. I have $\sum_{k=1}^n i^k = \frac{1}{2}(1-i)(-1+i^n)$ and now I get rid of $n$ how? I say that I choose $k$ arbitrary bigger than $n$? $\endgroup$ – Spaced Oct 20 '14 at 21:30
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I believe my problem is the argumentation. I have $\sum_{k=1}^n i^k = \frac{1}{2}(1-i)(-1+i^n)$ and now I get rid of $n$ how? I say that I choose $k$ arbitrary bigger than $n$?

No, you use the triangle inequality, $\lvert z\cdot w\rvert = \lvert z\rvert\cdot \lvert w\rvert$ and $\lvert i\rvert = 1$ to obtain

$$\left\lvert \sum_{k=1}^n i^k\right\rvert = \frac{1}{2}\lvert 1-i\rvert\cdot \lvert-1+i^n\rvert \leqslant \frac{1}{2}\sqrt{2}\cdot(\lvert -1\rvert + \lvert i\rvert) = \frac{1}{2}\sqrt{2}\cdot(1+1) \leqslant \sqrt{2},$$

independently of $n$.

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  • $\begingroup$ Thanks a lot, I will remember this for sure :-) $\endgroup$ – Spaced Oct 20 '14 at 21:37

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