# 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.

Last edited:

CRGreathouse
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