anonymous
  • anonymous
how can i prove that the series of sin(n) is divergent?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sum_{n=0}^{\infty}\sin(n)\]
anonymous
  • anonymous
any help please? tell me what are you thinking of
anonymous
  • anonymous
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.

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