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# The limit n --> infinity Sum(k=1 To n) | e^(2πik/n) - e^(2πi(k-1)/n)| is ...?

(A) 2;

(B) 2e;

(C) 2π;

(D) 2i;

Explain your answer...

### 1 Answer

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- Nick SLv 61 decade agoFavorite Answer
Since

| e^(2πik/n) - e^(2πi(k-1)/n)|

= | e^(2πik/n) | ⋅ | 1 - e^(-2πi/n)|

= | 1 - e^(-2πi/n) |, the sum is just

n | 1 - e^(-2πi/n) |.

Now e^(-2πi/n) = 1 - 2πi/n + O(1/n²).

Thus

n | 1 - e^(-2πi/n) | = n | 2πi/n + O(1/n²) |

= 2π + O(1/n).

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