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How would one proof if the complex series $s_N = \sum_{n=1}^N a_n$ with $a_n=i^n$ converges or diverges?

$s = \sum_{n=1}^\infty i^n$ where $i^2=-1$.

Thanks in advance for all tips, links and references!

PatrickSteiner
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  • Your notation makes no sense. $s_n$ has an $n$ appearing... $\sum\limits_{n=1}^\infty a_n$ will after evaluating not have any $n$ remaining. Do you mean to ask about $s_N = \sum\limits_{n=1}^N a_n$ instead? – JMoravitz Oct 29 '20 at 14:52
  • Then, to clarify, by $i$ here, you mean the complex number $i$ such that $i^2=-1$? Are you familiar with related results such as whether or not $t_N = \sum\limits_{n=1}^N (-1)^n$ converges or diverges? Note that $t_N$ follows the sequence $-1, 0, -1, 0, -1, 0,\dots$ – JMoravitz Oct 29 '20 at 14:54
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    As for proving the final result... there are surely dozens of ways, several of which rely on theorems which should have only been introduced after you have built some level of fundamental understanding. I strongly discourage using any of these shortcuts. Since you are asking this question (which should be one of the very first examples introduced) it is clear you need to build your fundamentals, you should approach directly by definition. What does it mean to converge? What does it mean to diverge? You should be able to find the $\epsilon$-$N$ definition and get comfortable using it. – JMoravitz Oct 29 '20 at 14:59
  • I am used to the Comparison-test, Alternating-series-test, Ratio-test, Root-test, etc. However I don't see how one should progress with one of these tests in this very case. – PatrickSteiner Oct 29 '20 at 15:08
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    I already answered to this question https://math.stackexchange.com/questions/3881652/showing-that-in-0-when-i-is-the-imaginary-unit-and-n-is-infinite/3881774#3881774 – Raffaele Oct 29 '20 at 15:43
  • Does this answer your question? Showing that $i^n=0$ when $i$ is the imaginary unit and $n$ is infinite – Raffaele Oct 29 '20 at 15:44
  • @Raffaele introducing Cesaro summation to a beginner is very dangerous. You will only confuse them. To emphasize... the series discussed here using normal summation is divergent. Periodt. The story can end there. The sequence of $i^n$ as $n$ approaches infinity is also divergent. $i^n=0$ is nonsense and should be dismissed as such. – JMoravitz Oct 29 '20 at 15:49
  • @Raffaele There is plenty with normal summation to confuse a beginning student. Riemann rearrangement theorem comes to mind. Wait until they have come to grips with normal summation and completed at least a course or two in real analysis before introducing Cesaro summation or other "supersums." Otherwise, we'll get more people confused asking about $-\frac{1}{12}$ who should never have been shown the result. – JMoravitz Oct 29 '20 at 15:55
  • @JMoravitz The other user actually tried to claim that the series converged to zero, so I told him that he could not mix methods for divergent series (Grandi's series) and methods for convergent ones. It was clear that the series diverges: I wrote it in the very first line – Raffaele Oct 29 '20 at 16:04

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A (real or complex) sequence $b_n$ is defined to "converge to a limit $b$" iff for every choice of $\epsilon>0$ there exists a choice of $N_0$ such that the implication $n>N_0 \implies |b_n - b|<\epsilon$ holds true.

By negating the condition:

A (real or complex) sequence $b_n$ is defined to "diverge" iff for every proposed limit $b$ there exists a choice of $\epsilon > 0$ such that for every choice of $N_0$ there exists some $n>N_0$ such that $|b_n-b|\geq \epsilon$

In lay terms, that is to say that a sequence converges if it gets closer and closer to a limit and that for any desired closeness there is some point in the sequence such that from that point on all terms in the sequence are at least that close.

On the other hand, a sequence diverges if regardless how far we look in the sequence there are always at least some terms which are farther away than we would like.

For your specific series, we look at the sequence of partial sums. We see that the sequence of partial sums goes $i,-1+i, -1, 0, i, -1+i, -1, 0, i, -1+i, -1, 0,\dots$ repeating this pattern ad infinitum.

We can see that by setting $\epsilon = \frac{1}{4}$ for instance that regardless what proposed limit we take, at least one of $0$ or $-1$ will be further away from the limit than $\frac{1}{4}$ and for any choice of $N_0$ there will always be some $n>N_0$ such that the term is equal to whichever of $0$ or $1$ is further than $\frac{1}{4}$ of the proposed limit.

Thus, the sequence diverges.

Note, the choice of $\frac{1}{4}$ here was entirely arbitrary. I could have used $\frac{1}{10}$ or $\frac{1}{\pi}$ or $0.42$ or whatever else so long as it was small enough. I just needed to pick something such that being within that range of one of the values in the sequence necessarily implied it was outside of that range of another value in the sequence.

As for taking shortcuts, I emphasize that given that you asked this question you should not be worrying about them yet. Focus on your basic definitions. That said, there are many ways you can see this series diverges very quickly. The most basic of which is the "basic divergence test."

You can prove that every series who converges must necessarily have the summands approach zero. The contrapositive of this statement is that if you have a series whose summands do not approach zero, then the series necessarily must diverge. Your summands here are all of absolute value equal to $1$ and do not approach zero.

JMoravitz
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