I'm learning mathematical analysis recently. My book gave the definition of $\pi$ as the limit of sequence $\left\{n \sin \frac{180^o}{n}\right\}$.
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 $$ \sin(n+1)t = \sin nt \cos t + \cos nt \sin t = \sin nt \cos t\left(1 + \frac{\tan t}{\tan nt}\right) \le \frac{n+1}{n} \sin nt $$ 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?
A simpler way.
Imagine a circle as an equilateral polygon with $n$ sides. And to make things easier, imagine the distance from polygon edges to the centre as the unit $1$ (kinda like a unit circle). Let's call it a "unit equilateral polygon".
Now, let's cut the polygon into $n$ small triangles like how you cut a cake.
Now, each of these triangles has two sides of unit $1$ and an angle of $\frac{360^\circ}{n}$ between them.
Let's cut this triangle itself into two right angle halves. The angle is now $\frac{180^\circ}{n}$. Notice the opposite of this angle is simply $\sin(\frac{180^\circ}{n})$
Multiply this by $2$ to get the length of one side of the polygon $2 \space \sin(\frac{180^\circ}{n})$.
Remember, there are $n$ of these sides. So
Circumference of n-equilateral polygon $= 2 \space n \space \sin(\frac{180^\circ}{n})$
Now, given a circle is nothing but an equilateral polygon with an infinite number of sides.
$\text{Unit Circle Circumference} = 2 \Pi = \lim_{n \to \infty} 2 \space n \space \sin(\frac{180^\circ}{n})$
we divide by $2$ to get
$\Pi = \lim_{n \to \infty} \space n \space \sin (\frac{180^\circ}{n})$
Job done.