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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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2 Answers 2

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


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: $$\lim_{u\to 0}{\left[\frac{\sin{u}}{u}\right]}=1$$



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

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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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