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Puzzled by sequence

  1. Jun 7, 2012 #1


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    So I was trying to show that

    [tex] a_n = n \tan\frac{\pi}{n} [/tex]

    is decreasing as n increases (n>=3) . I can't see it. Anyone may help ? :)
  2. jcsd
  3. Jun 7, 2012 #2
    tan x = x + 1/3x3 + 2/15x5 + 17/315x7 + .....for |x| < π/2, so expand an and subtract from it an+1 and you should see this
  4. Jun 7, 2012 #3


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    I was aiming at a method from first principles (tangent is the ratio of sine and cosine), without derivatives, because I know that I can take n from N (>=3) to the real subset [3,infty) and analyze the sign of the first derivative of the function. It's negative for x>0, hence the function keeps decreasing as x increases. Then the monotonicity of the sequence follows.

    Your definition of tangent needs to be proven equivalent to the triangle one.
  5. Jun 8, 2012 #4
    I tried the tangent difference identities and such but it does not help. I think it is best to use the sine-tangent inequality to evaluate the derivative of sine and cosine, and then using them to evaluate the derivative of tangent. I doubt there is any better way using elementary means.
  6. Jun 10, 2012 #5


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    Going further, the proof I'm searching reduces to show that once [itex] \displaystyle{n\in \mathbb{N}}[/itex], the sequence

    [tex] a_n = n \sin\frac{1}{n} [/tex]

    increases with n.

    How do I prove it ?
  7. Jun 10, 2012 #6
    I'm not sure about this, but here's what I thought.

    You can try seeing whether n increases faster than sin(1/n) decreases, by taking their derivatives(or some other method). You'll reach to the conclusion that n does increase at a faster rate than sin(1/n) decreases, hence, your original function increases.
  8. Jun 10, 2012 #7
    I think you did not understand his point. We can easily differentiate the function and conclude that it increases everywhere because it has a positive derivative. He wants a proof that includes only trigonometric identities, and I am sceptic about whether it is possible to present such a proof.

    I think that the first thing to do would be to express one of the sines in terms of the other one. To do this, I used this course of action:
    Then, I conclude that [itex]\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{1}{n}[/itex]. I now tried to apply this to the sine sum formula as follows:
    I then rearranged this into the inequality
    [tex]n\sin\left(\frac{1}{n+1}\right)\cos\left(\frac{1}{n(n+1)}\right)+n\cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n(n+1)}\right) \leq n\sin\left(\frac{1}{n+1}\right)+\sin\left(\frac{1}{n+1}\right)[/tex]
    which can be rewritten as
    [tex]n\cos\left(\frac{1}{n(n+1)}\right)+n\cot\left(1/(n+1)\right)\sin\left(\frac{1}{n(n+1)}\right)\leq n+1[/tex]
    It is straightforward to show that the first term in the LHS is less than or equal to n, so it appears we only need to prove
    [tex]\cot\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n(n+1)}\right)\leq 1/n[/tex]
    I am stuck here. This inequality is equivalent to the above function increasing monotonically.

    Can anyone carry this on?

    Note: After graphing the function using a program, I am convinced that the inequality holds when n is greater than one, and even more: 1/n and this function is asymptotic, their quotient has a limit of one.
    Last edited: Jun 10, 2012
  9. Jun 10, 2012 #8
    Here's what I did,

    We know that sin(x) < x, for all x.

    [tex]sin\left ( \frac{1}{n(n+1)} \right ) < \frac{1}{n(n+1)}[/tex]

    Substituting in the inequality Millennial had,

    [tex]cot\left ( \frac{1}{n+1} \right ) \cdot \frac{1}{n(n+1)} \leq \frac{1}{n} \\

    cot\left ( \frac{1}{n+1} \right ) \cdot \frac{1}{n+1} \leq 1[/tex]

    Having n>2 it follows that,

    [tex]tan\left ( \frac{1}{n+1} \right ) \cdot (n+1) \geq 1[/tex]

    We know that tan(x) > x for any x. Hence at the least possible condition, let us assume tan(x) = x. So we get,

    [tex]\left ( \frac{1}{n+1} \right )\cdot (n+1) \geq 1[/tex]

    Hence the proof.

    Please tell me if there is a gap in my logic, as this I've not much practice with such questions :smile:
  10. Jun 10, 2012 #9
    Yes, that is quite elegant. I did not see that, probably due to me being busy with other questions. Thanks for your help with my proof :)
  11. Jun 10, 2012 #10
    Yay!! I was solving some integrals too, this was a fun break :biggrin:

    Your method to get that inequality was really insightful. Probably wouldn't ever have thought of manipulating it that way. Thanks for the idea!
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