Seemingly contradictory results show $f(n) = n e^{\frac{-\pi n}{2} i }$ is divergent

Question

While I was doing an exercise in which I have to show that the sequence defined by

$$ f(n) = n e^{-\pi ni/2 }$$

is divergent, I found something I don't know how to explain : two different approaches give me two different results (both showing divergence) which seem contradictory.

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One way to show divergence ( in the sense that $f(n) \to \infty$) is to prove that $|f(n)| \to +\infty$ if $n \to +\infty$, so :

$$|f(n)| = |n e^{-\pi ni/2 }| = |n| |e^{-\pi ni/2 }| = n\sqrt{\cos^2(\frac{\pi n}{2})+\sin^2(\frac{\pi n}{2})} = n $$

which is enough to prove the above.

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But I can show divergence also proving that either the real part of $f(n)$ or the complex part doesn't converge.

First, I separate the function :

$$f(n) = n\cos(\frac{\pi n}{2})-in\sin(\frac{\pi n}{2})$$

and then I can say that since I can obtain $2$ subsequences on $\mathbb{N}$ going to different limits then the function diverges.

Actually, I can decompose $\mathbb{N}$ in $4$ subsequences covering all $\mathbb{N}$.

$$n_1 = 4k , n_2 = 4k+1, n_3 = 4k+2 , n_4 = 4k+3$$

Giving four different limits $$f(n_1) \to +\infty , f(n_2) \to i\cdot(-\infty),f(n_3) \to -\infty, f(n_4) \to i\cdot(+\infty) $$

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Even if I don't understand what $i \cdot \infty$ means, it seems to me that the two methods gives seemingly contradictory results because the first suggests that $f(n) \to +\infty$ if $n \to +\infty$, while the second suggests that the limits oscillates at infinity. How can this be explained?

Edit : Thanks to @Radost comment I see that the two main problems are :

• the definition of $i \cdot \pm \infty$
• the first method shows just that the magnitude of $f(n)$ tends to $+\infty$, therefore $f(n)$ tends to some type of infinity ( positive or negative or with the imaginary unit involved) but we don't know which one.

Answer 1

Since we are considering a sequence of complex numbers, we have to reconsider the meaning of the symbol $\infty$.

Whereas in the real line there are two "types" of infinity (the negative infinity $-\infty$ and the positive infinity $+\infty$), in the complex plane there is only one type of infinity, namely $\infty$, which represents the idea of a complex number "having infinite module".

Both of your proofs are correct and show the divergent character of the sequence. They don't contradict each other

Note that the module of $f(n)$ is $|f(n)|=n$, and its argument is $Arg(f(n)) = -\frac{\pi}{2}n$. We see two things:

1. The sequence $|f(n)|$ (of real numbers) goes to $+\infty$ as $n\rightarrow\infty$.
2. $Arg(f(n+4)) = Arg(f(n))$ for all $n=0,1,\dots$.

Notice how your proofs are not contradictory with each other. In the first one you have shown that the sequence has module which grows indefinitely, and the second one that you can extract four subsequences with different limits, which of course implies divergence of the initial sequence.

I hope this helps.

Answer 2

In your second method you saw the pattern of your sequence $f(n)$: from the step $n-1$ you achieve the step $n$ by increasing by a unit the distance from the origin and rotating the new point clockwise of 90 degrees. The $n$-point is going further from the origin while it is rotating. The definition of oscillating sequence is given for real sequence $g_n:\mathbb{N}\to \mathbb{R}$.

In case of complex sequences we define the convergence of $f(n)$ to a point $z\in \mathbb{C}$ if the distance $\lvert f(n)-z \rvert\to 0$ as $n\to \infty$. By definition the sequence $f(n)$ diverges if such $z$ does not exist, that is our case as you proved. The crucial point is that we need an order in the range (that is what $\mathbb{C}$ does not possess) to talk about oscillation.