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Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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 S
    Lv 6
    1 decade ago
    Favorite 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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