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Infinite series of tan(1/n)

  1. Nov 2, 2016 #1
    Question

    ∑ tan(1/n)
    n = 1
    Does the infinite series diverge or converge?

    Equations
    If limn → ∞ ≠ 0 then the series is divergent

    Attempt
    I tried using the limit test with sin(1/n)/cos(1/n) as n approaches infinity which I solved as sin(0)/cos(0) = 0/1 = 0

    This does not rule out anything and I cannot think of what else to try

    The answer stated that the series is divergent so I just need to know what test to use to determine that
     
  2. jcsd
  3. Nov 2, 2016 #2

    Mark44

    Staff: Mentor

    What other tests do you know about?

    Also, in future posts, please do not delete the Homework Template.
     
  4. Nov 2, 2016 #3
    Wait I think I got it

    Using the Limit Comparison Test with the infinite series of 1/n would give limn → ∞ tan(1/n)/(1/n) = n tan(1/n) = n sin(1/n)/cos(1/n)
    n sin(1/n) = sin(1/n)/(1/n) = 1 so the limit can be written limn → ∞ 1/cos(1/n) = 1/cos(0) = 1 so the limit = 1

    Since the limit is larger ≥ 0 that means that both series tan(1/n) and 1/n must converge or diverge and since 1/n obviously diverges tan(1/n) also diverges

    Please let me know if I made any mistakes and thank you
     
  5. Nov 2, 2016 #4

    Mark44

    Staff: Mentor

    Better:
    ##\lim_{n \to \infty}\frac{\tan(1/n)}{1/n} = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n} \frac 1 {\cos(1/n)} = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n}\cdot \lim_{n \to \infty} \frac 1 {\cos(1/n)} = 1 \cdot 1 = 1##
    Overall, looks good.
     
  6. Nov 3, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You can use a comparison test: for small ##x > 0## we have ##\sin x > x/2## and ##\cos x < 2##, so ##\tan x > x/4##, hence ##\tan(1/n) > 1/(4n)##.
     
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