# Definition of $\pi$, $\lim_{n \to \infty}{n \sin(180^o/n)}$

I'm learning mathematical analysis recently. My book gave the definition of $\pi$ as the limit of sequence $\{n \sin \frac{180^o}{n}\}$.

The way it prove this sequence is convergent is quite strange to me. It first showed the sequence is smaller than 4, then monotonically increasing.

The latter part is confusing. It first let $t = \frac{180^o}{n(n+1)}$, and proved $\tan nt \ge n\tan t$ for $nt \le 45^o$, so $$\sin(n+1)t = \sin nt \cos t + \cos nt \sin t = \sin nt \cos t(1 + \frac{\tan t}{\tan nt}) \le \frac{n+1}{n} \sin nt$$ then $$n \sin \frac{180^o}{n} \le (n+1) \sin \frac{180^o}{n+1}$$

This is perfectly correct, but how can I come up with a $t$ like this? If I'm to prove this, is there a way to figure out what the $t$ should be like? Or all I can do is just memorize it? Alternately, do you guys have a more intuitive proof?

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A more intuitive proof could look as follows:

1. Set $n=1/x$, then you have $\lim \limits_{x\to 0} \frac{\sin(ax)}{x}$.
2. Now do a series expansion to get $\lim \limits_{x\to 0} a-\frac{a^3x^3}{6} +\dots =a$
3. Plugin in $a=\pi$, if you like ...
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Here's a more intuitive proof: Take an n-sided polygon. Radius (center to vertex) R. Calculate the perimeter of the polygon. It will be:

P= 2*Rnsin theta, where theta is 360/2n (Get this by dropping a perpendicular from vertex to a side, calculate the length of the side, remembering that we've bisected theta and the side. multiply by n.)

Now watch this polygon become a circle: P=lim 2*Rnsin(180/n) = 2*(lim n*sin(180/n))*R. But it has to be 2*piR. So lim nsin(180/n), as n goes to infinity, must be pi.

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Here's a pretty simple proof I know for your problem:

$$\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}$$

Let $$\frac{a}{x}=u$$ $$\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}=\lim_{\frac{a}{u}\to\infty}{\left[\frac{a}{u}\cdot\sin{u}\right]}$$

With $$\frac{a}{u}\to\infty\Leftrightarrow u\rightarrow 0$$

$$\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}=a\cdot\lim_{u\to 0}{\left[\frac{\sin{u}}{u}\right]}$$

There is a theorem that says: $$\lim_{u\to 0}{\left[\frac{\sin{u}}{u}\right]}=1$$

$$\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{a}{x}}\right]}=a$$

q.e.d.

(Please note that $180°=\pi$. That's why $\Leftrightarrow\lim_{x\to\infty}{\left[x\cdot\sin{\frac{180°}{x}}\right]}=\pi$)

So notice that not any of the proofs here, not even yours, can be used to define pi, as a definition of something cannot contain that thing itself. It's like saying: "documentation is a word that means documentation"

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I disagree: His definition of $\pi$ is perfectly valid and non-circular, depending on how you define $\sin(x)$. If you define it using degrees, then this would give you a definition of $\pi$ perfectly well. – Simon Rose Oct 4 '14 at 20:27
Can't really agree, though... It's a matter of personal interpretation, I guess; according to me, 180° is exactly the same as pi rad, making the definition circular. I could very well be wrong, though... – mmhenrot Oct 4 '14 at 20:56