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Hurkyl's Sum

  1. Apr 26, 2003 #1

    Tom Mattson

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    Congratulations Hurkyl, you're famous.

    This was a thread that he started in PF v2.0, and which I participated in. I made a copy of it because I left it unfinished.



    My (as yet incomplete) solution:

    Parametrize the sum as follows:

    That gets rid of the nasty arctan function.

    My approach will be to find the sum of S'(a) and integrate with respect to 'a' with the limits 0<a<1. That will give me:
    S(1)-S(0)=S(1), which is the original sum.

    To be continued...
  2. jcsd
  3. Apr 26, 2003 #2


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    I had forgotten I put that here!

    That new sum still doesn't look pretty, but that doesn't mean it can't be summed! I'll be watching to see where you go from there!
  4. Apr 28, 2003 #3
    I think I have a solution...


    Let tan(a) = n+1; tan(b) = n;


    So arctan(1/(n^2+n+1))=arctan(n+1)-arctan(n)...
    Suming we obtain sum = arctan(infinity)-arctan(0) = pi/2...

    Is it good ?
  5. Apr 28, 2003 #4

    Tom Mattson

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    How did you jump from the first line to the second?
  6. Apr 28, 2003 #5


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    IT SURE IS!!! AWESOME, BRILLIANT solution, actually!!!

    My compliments, Dario

    The formula used is simply the one for the tangent of a sum(difference) of arcs
  7. Apr 28, 2003 #6

    Tom Mattson

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    Well, I guess I'm going to scrap my solution then.
  8. Apr 28, 2003 #7


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    Bah don't give up!

    I've seen at least one pretty solution that didn't use that cheap trick, it would be interesting to see another one!
  9. Apr 28, 2003 #8


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    The solution is actually good and brilliant since it uses a good hint of insight for the decomposition of the fraction and a pretty much unknown formula for addition of inverse trigonometric functions but if you want to give us your definition of "cheap trick" I am curious to read it!

    Said that, I am sure that Tom's approach can take somewhere at least the idea to focus on a derivate sum and integrate. I am not sure about the functional form he has chosen but it is a possibility...

    Last edited by a moderator: Apr 28, 2003
  10. Apr 28, 2003 #9


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    I was just teasing. :smile: I guess I didn't make it all that clear, though.

    I derived the sum via that very argument, and I present it as a challenge problem because I know how difficult it would be to spot.
  11. Apr 28, 2003 #10


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    gotcha ;)
  12. Apr 29, 2003 #11
    I hate using derivatives to "compute" sums...
  13. May 3, 2003 #12
    Bogdan's great solution remembered me that arctg[1/(n2+n+1)] can be convenably rewritten using some discrete math considerations and trigonometry also.

    If dx is a positive variation (not necessarilly infinitesimal) around a parameter x then we have:

    arctan(x+dx)-arctan(x)=arctan[dx/(1+x*dx+x2)] (1)

    where I used the trigonometric formula

    arctan(a)-arctan(b)=arctan[(a-b)/(1+a*b)] (2)

    If we replace in (1) x with n and put the condition that the step is dx=1 --->

    arctan(n+1)-arctan(n)=arctan[1/(1+n+n2)] (3)

    After simplifying the terms the initial sum can be rewritten as:

    S=limk->00∑;from n=0 to k arctan[1/(1+n+n2)]


    Of course I've written all these only in order to check some HTML mathematical symbols :-).

    [edit to correct some gramatical mistakes]
    Last edited: May 7, 2004
  14. May 3, 2003 #13
    arctan(dx/(1+x*dx+x^2))/dx=arctan(dx/(1+x*dx+x^2))/(dx/(1+x*dx+x^2))*1/(1+x*dx+x^2)=1*1/(1+x^2)....because lim arctanx/x=1, x->0...
    So...we obtain the "derivative" for arctan...
  15. May 4, 2003 #14

    I've just realized that you can also find the sum S of the series &#8721n=0&#8734 arctan[1/(n2+n+1)] by calculating the integral:

    I=&#8747x=0&#8734 arctan[1/(x2+x+1)]

    Indeed the function f(x)=arctan[1/(x2+x+1)] is continuous and decreasing over the interval [0,&#8734] and f(1)=term[1],f(2)=term[2] and so on.I is the seeked sum when n->&#8734.

    [edit to add]

    Of course I am wrong.At most the comparation with the above integral proves that the series is convergent...
    Last edited: May 4, 2003
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