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Homework Help: Converge or diverge and how

  1. Feb 22, 2007 #1
    1. The problem statement, all variables and given/known data

    sum of tan(1/n)/(1+n) for n=1 to infinity

    2. Relevant equations

    3. The attempt at a solution

    I tried using the ratio test and the comparison/limit comparison tests but can't think on anything to compare it to.
  2. jcsd
  3. Feb 22, 2007 #2
    I'm far from a math wiz, but here's how I see it:

    tan(x) will always be between -1 and 1,

    but tan (1/n) will always be positive so it will be between 0 and 1

    (well, actually for this problem, the max value of the numerator is tan(1) and it continues to decrease)

    and the denominator will be at the very least 2 and it will continue to increase

    so your fraction will continue to get smaller and smaller

    in fact, it will always be smaller than 1/n

    since the top will be at most 1 and the the denominator will be greater than n

    so, tan(1/n)/(n+1) < 1/n

    and you know that 1/n converges...so by comparison, tan(1/n)/(1+n) must also converge
    Last edited: Feb 22, 2007
  4. Feb 23, 2007 #3
    Are you sure about that? What does tan(x) approach as x approaches pi/2 from the left?
  5. Feb 23, 2007 #4
    The SEQUENCE with general term 1/n converges, does the series?
  6. Feb 23, 2007 #5
    Well.. sin(1/n) is approximately 1/n for large n. I would do some sort of error estimate for tan(1/n) = 1/n + error to see if the the series CONVERGES absolutely. (Error estimates for sin(1/n) may suffice.)

    Watch out for comparing to the famously DIVERGING harmonic series.
  7. Feb 23, 2007 #6
    d_leet, ok tan(x) does approach infinity when cos(x) approaches 0


    "The harmonic series diverges, albeit slowly, to infinity"


    well...ok, I forgot about the n^r r > 1 condition for convergence

    but that's why I put the disclaimer about not being a math wiz :-p

    but...BUT...the series in question does converge

    <---has super maple skills
  8. Feb 23, 2007 #7
    Thanks so much but doesnt 1/n diverge by p-series because p=1. remember p-series diverges where 1/(n^p) p<=1
  9. Feb 23, 2007 #8
    mybsaccownt sorry about the last post i didn't fully read your last one.
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