32 views

The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is

1. $2$
2. $2e$
3. $2 \pi$
4. $2i$
in Calculus
retagged | 32 views

+1 vote

Answer: $C$

The given expression can be simplified as:

$$=\bigg| e^{\frac{2\pi ik}{n}}- e^{\frac{2\pi i(k-1)}{n}}\bigg|$$

$$= \bigg| e^{\frac{2\pi i(k-1)}{n}}\bigg |\bigg |e^{\frac{2\pi i}{n}}-1\bigg |$$

$$= \bigg|\cos\bigg (\frac{2\pi}{n}\bigg )-1 + i\sin\bigg (\frac{2\pi}{n}\bigg )\bigg |$$

$$= \bigg | -2\sin^2\bigg (\frac{\pi}{n}\bigg) + 2i\sin\bigg(\frac{\pi}{n}\bigg)\cos\bigg (\frac{\pi}{n}\bigg)\bigg |$$

$$= 2\sin\bigg(\frac{\pi}{n}\bigg )$$

Now, the required limit will be:

$$\therefore \underset{n \to \infty }{\lim} \sum^{n}_{k = 1}2\sin \bigg (\frac{\pi}{n}\bigg ) = \underset {n \to \infty}{\lim}2n\sin\bigg (\frac{\pi}{n}\bigg ) = 2\pi$$

$\therefore$ C is the right answer.

by Boss (19.1k points)
edited by
0

Here's a mistake in the line shown below.

Besides can't you make some side notes like below?

$|x+iy|=\sqrt{x^2+y^2}$
$e^{i\theta}=\cos{\theta}+i\sin{\theta}\Rightarrow |e^{i\theta}|=\sqrt{\cos^2\theta+\sin^2\theta}=1$

0
thanks.

Side notes like?

You mean more explanation?
+1
Yeah. But please correct the mistake first. Side notes are optional although it helps others to know understand thoroughly. 🙂
+1
done.

+1 vote