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I notice that,

$i^n = i, \quad if \text{ } n \equiv \text{ } 1 \text{ } mod \text{ } 4$

$i^n = -1, \quad if \text{ }n \equiv \text{ } 2 \text{ } mod \text{ } 4$

$i^n = -i, \quad if \text{ }n \equiv \text{ } 3 \text{ } mod \text{ } 4$

$i^n = 1, \quad if \text{ } n \equiv \text{ } 0 \text{ } mod \text{ } 4$

so the series $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$ seems to be an alternating series that does not converge to any point in the complex plane. Is there a rigorous way to prove this divergence? How would you write a proof to this divergence or convergence (if I am wrong)?

Thanks!

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    $\begingroup$ Show that the modulus of the $n$-the term does not tend to $0$ $\endgroup$ – Ant Jan 28 '17 at 20:23
  • $\begingroup$ What you should write always depends on what you can use in the proof. But the point is, of course, that you first prove that the term doesn't tend to 0, and then use or prove the general fact that in this case the series diverges. $\endgroup$ – Wolfram Jan 28 '17 at 20:28
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Since $\forall n\in\mathbb{N},\,\left|2+i^n\right|\in\{1,\sqrt 5,3\}$, we do not have $\lim_{n\to\infty}\left|\frac{1}{2+i^n}\right|=0$ (in fact, that limit barely doesn't exist !).

So, the series diverges.

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  • $\begingroup$ Not sure what you mean by "in fact, that limit barely doesn't exist " $\endgroup$ – zhw. Jan 28 '17 at 21:38
  • $\begingroup$ For any sequence converging to some real $L$, we know that every subsequence converges also to $L$. But in that case, we have got three subsequences, namely $u_{4n}$, $u_{4n+2}$ and $u_{4n+1}$, which converges to 3 distincts values. $\endgroup$ – Adren Jan 28 '17 at 21:41
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For $n\geq 0$, let $$u_n=\frac{1}{2+i^n}$$ and

$$S_n=\sum_{k=1}^n u_k.$$

we have

$$S_{4n+1}-S_{4n}=\frac{1}{2+i}$$

thus, $(S_n)$ is not a Cauchy sequence and doesn't converge.

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