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Series question

  1. Jan 16, 2009 #1
    I have this series [tex]\sum_{k=1}^{\infty}\frac{\arcsin{\displaystyle\frac{1}{k}}}{k^2}[/tex]

    I intended to prove this by Direct Comparison Test using the inequality:
    [tex]\frac{\arcsin{\displaystyle\frac{1}{k}}}{k^2} \leq \frac{1}{k^2}[/tex]

    Can I justify this inequality by saying that [tex]\arcsin{\frac{1}{x}}[/tex] is a decreasing function?
  2. jcsd
  3. Jan 16, 2009 #2


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    IF you can show for some value of k that arcsin(1/k)< 1, then for that k
    [tex]\frac{arcsin \frac{1}{k}}{k^2}< \frac{1}{k^2}[/tex]
    and so for k larger, arcsin(1/x) is still less than 1 and
    [tex]\frac{arcsin \frac{1}{k}}{k^2}< \frac{1}{k^2}[/tex].
    Last edited by a moderator: Jan 16, 2009
  4. Jan 16, 2009 #3
    That inequality wouldn't quite hold, take k=1, for example. But in any case, this also wouldnt make any difference, since we know by comparison test that if a_n<b_n holds for some n>N, then if the series sum(b_n) converges so does sum(a_n).
    So, that inequality holds for k>N, you can figure out what N is in this case, which would also prove it, since adding or subtracting a finite number of terms from a series does not affect its convergence.

    Now, we know that sin(x) is defined for any x, however, its inverse(s) are defined on intervals with length pi, usually when we speak of the inverse of sin, we speak of its inverse in the interval


    and since sin(x) itself is upper bounded by 1, and lowerbounded by -1, it means that the dom. of its inverse will be [-1,1], while on the other hand it will be upperbounded by pi/2.

    So, if we wanted that inequality to hold for any k, then i think it should look sth like this, which at the end doesn't really matter as far as the conv/div of the series in question is concerned:

    [tex]\frac{\arcsin{\displaystyle\frac{1}{k}}}{k^2} \leq \frac{\pi}{2}*\frac{1}{k^2}[/tex]

    edit:Halls was faser :approve:
  5. Jan 16, 2009 #4
    Why don't you use the integral test...?

    [tex]\int_{1}^{\infty} \frac{\arcsin{\frac{1}{k}}}{k^2} dk \Rightarrow -\int \arcsin{u} du = -\left(\frac{1}{k} \arcsin{\frac{1}{k}} + \sqrt{1-\frac{1}{k^2}}\right)_{1}^{\infty} = -(0-\frac{\pi}{2} + 1 - 0) = \frac{\pi}{2} - 1 [/tex]

    Assuming my math is correct...
    Last edited: Jan 16, 2009
  6. Jan 16, 2009 #5
    So I obviously fail at spoilers but that should work I believe.
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