# Meta-mathematical Pursuits - Sine(Infinity)

I have noticed that, in the theory of divergent series and divergent integrals, $$\sin ( \infty ) = 0$$.
For example
$\sum_{n=0}^\infty \ (-1)^n = 1 - 1 + 1 - 1 ... = \frac{1}{1-(-1)} = \frac{1}{2}$

but

$\sum_{n=0}^k \ (-1)^n = \frac{ - (-1)^{n} - 1}{1-(-1)}[/tex] and taking the limit as $$k \rightarrow \infty$$ (use Euler's identity to put the negative power in trigonometric form) $$- \lim_{\substack{k \rightarrow \infty}} \frac{(\cos( \pi k ) + i * \sin ( \pi k )) - 1}{2} = 1/2$$ and hence that [itex]\sin ( \infty ) = \cos ( \infty ) = 0$ (Equate real and imaginary parts).

What I would like to know is why this theory (that sine is zero at infinity) is not generally accepted even though it allows for such things as the solution to
$$\lim_{\substack{x \rightarrow \infty}} \frac{\sin ( x ) + x}{x}$$
by use of l'Hospital's rule. Is there a good reason or is it simply too good of an example of a function that diverges by oscillation?

Please forgive any mistakes as this is the first time I've used LaTEX.

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## Answers and Replies

CRGreathouse
Science Advisor
Homework Helper
is it simply a too good of an example of a function that diverges by oscillation?

Pretty much. It has an essential singularity at infinity.

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
What I would like to know is why this theory in not generaly accepted
I'm not sure what you mean by this. I can take a guess....

Most people simply don't have a need to study esoteric functionals -- for the most part, it would do far more harm than good to try and teach them to people who don't have the background to understand what's going on, and when they would be useful.

(Contrast with things like distributions -- the dirac delta is so useful that it does more good than harm to teach it to people who can't yet understand what's going on)