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anonymous

  • 5 years ago

how can i prove that the series of sin(n) is divergent?

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  1. anonymous
    • 5 years ago
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    \[\sum_{n=0}^{\infty}\sin(n)\]

  2. anonymous
    • 5 years ago
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    any help please? tell me what are you thinking of

  3. anonymous
    • 5 years ago
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    You can write the taylor series for sin(n) and use the ratio test. \[\sum_{n=0}^{\infty}(-1)^n*x ^{2n+1}/(2n+1)!\] But you want to change the series index to n = 1 to perform the ratio test. You should get an infinite limit, which shows that the series diverges.

  4. anonymous
    • 5 years ago
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    if i use the ratio test, it will be convergent for all values of x actually i didn't understand your point

  5. anonymous
    • 5 years ago
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    Yeah, I see we want to sum all the sin(n)'s, not the sum of the terms of the representation of sin(n)...

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