Section 1.4 Example 8

[Min Gyu Woo]

Can someone prove this using the definition of limits for sequences:

$\lim_{n\rightarrow\infty} (1/n)(\cos{(n\pi/4)}+i\sin{(n\pi/4)}) = 0$

[Alexander Elzenaar]

We simply need to show that the absolute value of $(1/n)\bigl(\cos(n\pi/4) + i\sin(n\pi/4)\bigr) $ tends to zero as $ n \to \infty $. Note first that $ \lvert (1/n)\bigl(\cos(n\pi/4) + i\sin(n\pi/4)\bigr)\rvert = 1/n $ (the trig functions vanish due to the Pythagorean identity).

I claim that for every $\varepsilon > 0 $ there exists some $ N \in \mathbb{N} $ such that for all $ n > N $ we have $ \lvert 1/n \rvert < \varepsilon $. Let such a $ \varepsilon $ be given; then by the Archimedian property of the real numbers, there exists a natural number $ N $ such that $ 1/\varepsilon < N $. If we pick any $ n > N $, then $ 1/\varepsilon < n $, and $ 1/n < \varepsilon $; we are done. (Note: I am justified in dropping the absolute value bars because everything is positive.)

[Min Gyu Woo]

Could you explain more about the trig function vanishing?

[Victor Ivrii]

Could you explain more about the trig function vanishing?

" trig function vanishing" is the product of your imagination. Essentially you are told that
$$
| \frac{1}{n}\bigl(\cos (\theta)+i\sin(\theta)\bigr| = \frac{1}{n}  |\cos (\theta)+i\sin(\theta|= \frac{1}{n}.
$$

[Vedant Shah]

Can we use squeeze theorem on $\cos{\frac{n \pi}{4}}$ and $ i \sin{\frac{n \pi}{4}}$?
Then the sum of the limits is the limit of the sums (provided they both exist)

[Victor Ivrii]

Explain, how you use squeeze theorem.

[Vedant Shah]

Start with:
$-1 \le \sin(\frac{n \pi}{4}) \le 1$
$\frac{-1}{n} \le \frac{\sin(\frac{n \pi}{4})}{n} \le \frac{1}{n} $
$\lim_{n \to \infty} \frac{-1}{n} \le \lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le \lim_{n \to \infty} \frac{1}{n}$
$ 0 \le lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le 0$

Thus by squeeze theorem,
$\lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} = 0$

After a similar argument, we get that:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0$

We also have:
$lim_{n \to \infty} i = i$

Since all three limits exist, we have:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + i \frac{sin(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + lim_{n \to \infty} i *lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0+i0 = 0$

[Victor Ivrii]

Yes, it is an overcomplicated proof.

"mathoperators" like lim, sin, cos , ... should be typed as \lim, \sin, \cos ... to produce a proper output