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Find the sum of arctan(1/2n^2)

  1. Jun 24, 2005 #1
    [tex]\sum_{i=1}^{n} \arctan(\frac{1}{2n^2})[/tex]

    I have to find sum of first n terms.
    I have no idea what to do with these arctan. Give me please any hint.
     
  2. jcsd
  3. Jun 24, 2005 #2

    dextercioby

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    The answer is immediate.

    [tex] \sum_{i=1}^{n} \arctan\left(\frac{1}{2n^{2}}\right) =n\arctan\left(\frac{1}{2n^{2}}\right) [/tex]

    Daniel.
     
  4. Jun 24, 2005 #3
    I got that
    [tex]S_n=\arctan\frac{\tan(S_{n-1})+1/2n^2}{1-\frac{\tan(S_{n-1})}{2n^2}}[/tex]
    [tex]S_n[/tex] is sum of first n terms.
    But i'm not sure if it could help.
    I just started studying series by myself. Could you elaborate your solution, Dexter. I don't understand
     
  5. Jun 24, 2005 #4

    dextercioby

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    I've already given you the answer.Yours is not the answer to the problem written in post #1.

    Daniel.
     
  6. Jun 24, 2005 #5
    Oh, yes. It's my mistake.
    [tex]\sum_{n=1}^{\infty} \arctan(\frac{1}{2n^2})[/tex] is supposed to be. Sorry
     
  7. Jun 24, 2005 #6

    dextercioby

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    I haven't got a clue so far to prove that

    [tex] \sum_{n=1}^{\infty} \arctan\left(\frac{1}{2n^{2}}\right) =\frac{\pi}{4} [/tex]

    Daniel.
     
  8. Jun 24, 2005 #7
    Integration Daniel?
     
  9. Jun 24, 2005 #8

    dextercioby

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    Integrate what...?I don't see a pattern to form a Riemann sum.

    Daniel.
     
  10. Jun 24, 2005 #9
    I tried it out, no light.Where did you get thid problem from?
     
  11. Jun 24, 2005 #10

    Hurkyl

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    Wee, this is a variant of my favorite problem! The way I know how to do it involves the application of a trig identity...
     
    Last edited: Jun 24, 2005
  12. Jun 24, 2005 #11
    Hm. Unfortunately it's too subtle hint for me.
     
  13. Jun 24, 2005 #12
    [tex]\frac{1}{2n^2} = \frac{2}{4n^2}[/tex] u can use this
    to get
    [tex]arctan(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}[/tex]
    You can use the identity for inverse
    and try to figure out rest
     
  14. Jun 24, 2005 #13
    ok just saw ur post::
    arctana+arctanb = arctan(a+b)/(1-ab)

    so u will have arctan{1/22}= arctan(2n+1)-arctan(2n-1)

    now u can substitute for n=1,2,3,......k u will finally have

    arctan{1/n2}= arctan(2k+1)-arctan1
    now as k=> inf u will have pi/2-pi/4 = pi/4


    "what is here may be elsewhere, what is not here is nowhere."
    Prayers are empowered by a clear mind and heart, and purity of intention. With clarity, our consciousness aligns with higher consciousness, where the forces of nature and laws of creation operate according to divine order
     
    Last edited: Jun 24, 2005
  15. Jun 24, 2005 #14

    Hurkyl

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    Well, I was trying not to actually do his homework for him. :tongue2:
     
  16. Jun 24, 2005 #15
    At any rate, it is a very neat problem. I had not seen it before. Thanks to the OP for posting it. I wonder if there are other ways to do it ...?
     
  17. Jun 25, 2005 #16
    [tex]\arctan{1/2n^2}= \arctan(2n+1)-\arctan(2n-1)[/tex]
    Yes, it's very tricky!
    I also got that [tex]S_n=\arctan(\frac{n}{n+1})[/tex] (This was my question)
    This also leads to pi/4 when n->infinity.
    Thank you very much to all.
     
    Last edited: Jun 25, 2005
  18. Jun 25, 2005 #17

    saltydog

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    Thanks guys. Very clever. This is a summary which I take no credit for:

    Using Himanshu's technique above:

    [tex]\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})[/tex]

    Let:

    [tex]\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}[/tex]

    Now,

    [tex]Arctan(a)-Arctan(b)=Arctan\left[\frac{a-b}{1+ab}\right][/tex]

    Letting:

    a=(2n+1)
    b=(2n-1)

    We have:

    [tex]Arctan[\frac{1}{2n^2}]=Arctan\left[\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right]=Arctan(2n+1)-Arctan(2n-1)[/tex]

    Thus we have:

    [tex]\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})=\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right][/tex]

    So that using a(x) and a(y) to save space we have:

    [tex]
    \sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]=[/tex]

    [tex](a3-a1)+(a5-a3)+(a7-a5)+(a9-a7)+ . . .+[a(2n+1)-a(2n-1)]+ . . .
    [/tex]

    Noting that all but -a1 and the a(2n+1) terms cancel, we take the limit of this sum and find we are left with:

    [tex]\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]\to a(\infty)-a(1)[/tex]

    Since:

    [tex] Arctan(1)=\frac{\pi}{4}[/tex]

    and:

    [tex]\mathop\lim\limits_{n\to\infty}Arctan(n)=\frac{\pi}{2}[/tex]

    we finally obtain:

    [tex]\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}[/tex]

    That is beautiful!
     
    Last edited: Jun 25, 2005
  19. Jun 25, 2005 #18

    saltydog

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    Oh yea, I almost forgot. What about these:

    [tex]\sum_{n=1}^{\infty}ArcSin[\frac{1}{2n^2}][/tex]

    [tex]\sum_{n=1}^{\infty}ArcCos[\frac{1}{2n^2}][/tex]

    [tex]\sum_{n=1}^{\infty}ArcCot[\frac{1}{2n^2}][/tex]

    I assume there must be an equivalent relation for the ArcCot. I'll look into it.
    Wait a minute, isn't the cot=1/tan so can I propose that last one is just [itex]\frac{4}{\pi}[/itex]?
     
  20. Jun 25, 2005 #19

    Hurkyl

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    A sum of reciprocals is generally not the reciprocal of the sum.

    One of the advantages of the tangent function is that its sum formula involves only tangents, so the corresponding formula for arctangents is nicer.

    You could try constructing similar things for the others, I suppose. Start with a telescoping series, then apply an addition formula and see what you get.
     
  21. Jun 25, 2005 #20

    saltydog

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    Hello Hurkyl. You got to it before I could make amends. Yea, ArcCot and ArcCos will not converge. However, I suspect the ArcSin will. :smile:

    You know, it's interesting just looking at the functions. Note that ArcTan and ArcSin are 0 at 0 where the other 2 are >0. Thus intutitively one would suspect taking an infinite sum of a function closer and closer to x=0 for functions which are >0 would not converge whereas if the functions are 0 at x=0, the sum I suspect could converge. Suppose that's all embodied in the various tests for convergence. :redface:
     
    Last edited: Jun 25, 2005
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