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Why is cos(1/n) an increasing function?

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data
    I have the series an = (-1)^{n} cos(1/n) and I have to determine whether it converges or diverges.

    2. Relevant equations

    I used the Leibniz criterion

    3. The attempt at a solution
    However, I determined that bn = cos(1/n) is a decreasing function because:

    n+1 > n

    1/(n+1) < 1/n

    cos(1/(n+1)) < cos(1/n)

    ie cos(1/(n+1)) - cos(1/n) < 0

    so bn+1 - bn is < 0 meaning it's decreasing no?

    I said it diverged because lim n -> ∞ bn = 1 (and not 0 which is the condition for it to converge)

    I got the answer correct, but for the wrong reasons...

    How do you show that cos(1/n) is an increasing function? Or any function for that matter? I thought my n+1 > n method was valid most (if not all) functions. Obviously not :(

    Thanks a lot!
     
    Last edited: Jun 1, 2013
  2. jcsd
  3. Jun 1, 2013 #2

    Mentallic

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    Is this true? Think about the value of cos(x) for x approaching zero.
     
  4. Jun 1, 2013 #3
    I don't know whether it's true or not, that's the thing
    cos(x) x approaching 0 is 1 right?
     
  5. Jun 1, 2013 #4
    [cos(1/n)]' = - sin(1/n) / n^2

    ie f'(n) < 0 for all n because of the negative sign, so it should be decreasing?
     
  6. Jun 1, 2013 #5
    ok I understood

    n+1 > n

    1/(n+1) < 1/n

    sin(1/(n+1)) < sin(1/n)

    ie sin(1/(n+1)) - sin(1/n) < 0

    BUT if it's cos:

    n+1 > n

    1/(n+1) < 1/n

    cos(1/(n+1)) > cos(1/n)

    ie cos(1/(n+1)) - cos(1/n) > 0

    gonna learn it like:

    cos of something with a larger denominator is > cos of something with smaller denominator and
    sin of something with a larger denominator is < sin of something with smaller denominator
     
  7. Jun 1, 2013 #6

    Mentallic

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    Yes, cos(x) is decreasing for 0<x<pi/2 which means that if we take a positive integer n, then 1/n > 1/(n+1) and as such, because cos(x) is decreasing, cos(1/n) < cos(1/(n+1))
     
  8. Jun 1, 2013 #7
    Thanks!
     
  9. Jun 1, 2013 #8

    Mentallic

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    You're welcome! And by the way,

    What I wanted you to think about is that as x approaches 0 (thus getting smaller and smaller) then the value of cos(x) approaches 1 (thus getting bigger and bigger).
     
  10. Jun 1, 2013 #9

    Mute

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    Just in case it's not entirely clear to you, when

    x < y,

    it follows

    f(x) < f(y)

    only if x and y are both contained in a range for which f(x) is a monotonically increasing function: f'(x) > 0.

    If the function is monotonically decreasing (f'(x) < 0) on a range for which both x and y are in, then

    f(x) > f(y).

    This second case corresponds to the problem here.

    Lastly, note that these are not the only possibilities. If f(x) is not monotonic on a range that contains both x and y, you can get any of f(x) < f(y), f(x) > f(y) or f(x) = f(y) for x < y. For example, if ##y = x + 2\pi##, then x < y, but for f(x) = cos(x), f(x) = f(y).
     
  11. Jun 1, 2013 #10
    ##\frac{d}{dn} \cos{(\frac{1}{n})} = - \sin{(\frac{1}{n})} \cdot (- \frac{1}{n^2})##

    You missed a minus sign.
     
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