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watchmath

  • 5 years ago

Determine the convergence of the series \[\sum_{n=1}^\infty 1-\cos(1/n)\] (this is a problem from one of the student here)

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  1. anonymous
    • 5 years ago
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    is this open for anyone to answer?

  2. watchmath
    • 5 years ago
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    sure

  3. anonymous
    • 5 years ago
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    i expanded cos(1/n) and rewrote the summand as: \[\sum_{n=1}^{\infty}(-1)^{n+1}/(2n)!n^{2n}\] which is an alternating series that is monotone decreasing, so it converges.

  4. watchmath
    • 5 years ago
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    you mean using the Taylor series? then shouldn't we have to sigma notation?

  5. watchmath
    • 5 years ago
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    two sigma notations

  6. anonymous
    • 5 years ago
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    oh yep i made a mistake.

  7. anonymous
    • 5 years ago
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    hey ty for posting it up the question but whats the answer is ∑n=1∞(−1)n+1/(2n)!n2n

  8. anonymous
    • 5 years ago
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    posted by rsvitale

  9. watchmath
    • 5 years ago
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    no, he made a mistake. I'll post the solution if nobody answer this :).

  10. toxicsugar22
    • 5 years ago
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    hi watchmath can u help me after him

  11. anonymous
    • 5 years ago
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    okay lol

  12. toxicsugar22
    • 5 years ago
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    watchmath can u help me

  13. anonymous
    • 5 years ago
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    by the way do you know how to find the general term of a series

  14. anonymous
    • 5 years ago
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    if given something like this 3,6,10,15

  15. anonymous
    • 5 years ago
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    i mean sequence

  16. watchmath
    • 5 years ago
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    you need to try to see the pattern and formulate it in general.

  17. anonymous
    • 5 years ago
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    so you basically guess

  18. watchmath
    • 5 years ago
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    No guessing is not the right word for it. It is observing some regularity :).

  19. watchmath
    • 5 years ago
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    Ok, let solve the series problem. By the double angle formula we know that 1- cos(2x)= 2sin^2(x) It follows that \(1-\cos(1/n)=2\sin^2(1/(2n))\) Recall that for positive x we have \(\sin x < x\) It follows that \(\sin^2(1/(2n))< (1/(2n) )^2\) But know the series \(\sum 1/(2n)^2\) is convergent (p-series with p=2). Hence \(\sum 2\sin^2(1/(2n))\) is convergent as well :).

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