An explanation for the result of the following limit

Question

Whilst having troubles in calculating the following limit, which I thought it were indeterminate, I decided to put it into W. Mathematica (the serious software, not W. Alpha online) and it returned something weird. Could someone please explain me what is this?

$$\lim_{n\to+\infty} \frac{(-1)^n}{\sin(n)}$$

W. Mathematica's output:

$$e^{2 i \text{Interval}[\{0,\pi \}]} \text{Interval}[\{-\infty ,-1\},\{1,\infty \}]$$

Furthermore

A proof about the non existence of the limit would be much appreciated!

Answer 1

I wonder if we could give a very simplified but effective proof of the nonexistence of the limit $$\lim_{n\to+\infty} \frac{1}{\sin(n)}$$

Suppose it does exist the limit $\lim_{n\to+\infty} \sin(n) = L\neq 0$. Then:

$$\lim_{n\to+\infty} \frac{1}{\sin(n+1)} = \lim_{n\to+\infty} \frac{1}{\sin(n)\cos(1) + \sin(1)\cos(n)} = \frac{1}{L\cos(1) + \sin(1)\cos(n)}$$

Then suppose the limit $\lim_{n\to+\infty} \cos(n) = M\neq 0$ exists too. Then

$$\lim_{n\to+\infty} \frac{1}{\sin(n+1)} = \frac{1}{L\cos(1) + M\sin(1)}$$

Now we do the same thing for the cosine:

$$\lim_{n\to+\infty} \frac{1}{\cos(n+1)} = \lim_{n\to+\infty} \frac{1}{\cos(1)\cos(n) - \sin(1)\sin(n)} = \frac{1}{M\cos(1) - L\sin(1)}$$

If $n\to+\infty$ then $n+1 \sin n$, so we could write

$$\lim_{n\to+\infty} \frac{1}{\sin(n+1)} = \frac{1}{L} ~~~~~~~ \lim_{n\to+\infty} \frac{1}{\cos(n+1)} = \frac{1}{M}$$

In this optics

$$\frac{1}{L} = \frac{1}{L\cos(1) + M\sin(1)}$$

$$\frac{1}{M} = \frac{1}{M\cos(1) - L\sin(1)}$$

Now let's take the reciprocals

$$L = L\cos(1) + M\sin(1)$$ $$M = M\cos(1) - L\sin(1)$$

Multiplying the first one by $L$ and the second one by $M$, adding them we find

$$1 = \cos(1)$$

Which is a contradiction.

Answer 2

If you prove that $\lim_{n \rightarrow +\infty}\frac{1}{\sin^{2}(n)}$ does not exists then you have proved your result because if $\lim_{n \rightarrow +\infty}\frac{-1^{n}}{\sin(n)}$ exists then we have that $\lim_{n \rightarrow +\infty}\frac{1}{\sin^{2}(n)}=\lim_{n \rightarrow +\infty}\frac{-1^{2n}}{\sin^{2}(n)}=\lim_{n \rightarrow +\infty}\frac{-1^{n}}{\sin(n)} \cdot \lim_{n \rightarrow +\infty}\frac{-1^{n}}{\sin(n)}$ exists and we have a contradiction.

But I don´t know how to prove that $\lim_{n \rightarrow +\infty}\frac{1}{\sin^{2}(n)}$ does not exists.
I don´t know if $\lim_{n \rightarrow +\infty}\frac{1}{\sin^{2}(n)}$ does not exists.