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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. $2i$
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2 Answers

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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.

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Answer: C


Explanation:

Note that $$e^{\frac{2\pi \iota k}{n}}\enspace ;1\le k\le n$$ can be thought of as a roots of unity.

Since roots of unity form a regular n-gon. Hence, the side length of the side of n-gon is given by $$|e^{\frac{2\pi \iota k}{n}}-e^{\frac{2\pi \iota (k-1)}{n}}|$$.

Thus, the sum $$\sum_{k=1}^n |e^{\frac{2\pi \iota k}{n}}-e^{\frac{2\pi \iota (k-1)}{n}}|$$ is the perimeter of the regular n-gon.

 

Further, one should note that as $n\to \infty$, the regular n-gon forms a circle.

Thus, $$\lim_{n\to \infty} \sum_{k=1}^n |e^{\frac{2\pi \iota k}{n}}-e^{\frac{2\pi \iota (k-1)}{n}}|=2\pi$$

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