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Posers and Puzzles

Posers and Puzzles

  1. 25 Mar '08 17:36
    See the following interesting 3 series:

    (1*2)/(1*3) + (1*2*3)/(1*3*5) + (1*2*3*4)/(1*3*5*7) + (1*2*3*4*5)/(1*3*5*7*9) + ... = pi/2

    (1*2)/(3*5) + (1*2*3)/(3*5*7) + (1*2*3*4)/(3*5*7*9) + (1*2*3*4*5)/(3*5*7*9*11) + ... = pi/2 - 4/3

    (1*2)/(5*7) + (1*2*3)/(5*7*9) + (1*2*3*4)/(5*7*9*11) + (1*2*3*4*5)/(5*7*9*11*13) + ... = -3pi/2 + 24/5

    Which can be written also like this:

    (1)/(1) + (1*2)/(1*3) + (1*2*3)/(1*3*5) + (1*2*3*4)/(1*3*5*7) + (1*2*3*4*5)/(1*3*5*7*9) + ... = pi/2 + 1

    (1)/(3) + (1*2)/(3*5) + (1*2*3)/(3*5*7) + (1*2*3*4)/(3*5*7*9) + (1*2*3*4*5)/(3*5*7*9*11) + ... = pi/2 - 1

    (1)/(5) + (1*2)/(5*7) + (1*2*3)/(5*7*9) + (1*2*3*4)/(5*7*9*11) + (1*2*3*4*5)/(5*7*9*11*13) + ... = -3pi/2 + 5

    Another way to write the third one:

    (1)/(3) + (1*1)/(3*5) + (1*1*2)/(3*5*7) + (1*1*2*3)/(3*5*7*9) + (1*1*2*3*4)/(3*5*7*9*11) + ... = -pi/2 + 2

    My question is, can this "sequence of series" be continued, and the sum is always of the form a*pi+b with a,b rational?

    (1*2)/(7*9) + (1*2*3)/(7*9*11) + (1*2*3*4)/(7*9*11*13) + (1*2*3*4*5)/(7*9*11*13*15) + ... = ?

    (1*2)/(9*11) + (1*2*3)/(9*11*13) + (1*2*3*4)/(9*11*13*15) + (1*2*3*4*5)/(9*11*13*15*17) + ... = ?

    Maybe someone has Maple/Mathematica or a similar software which can evalute these series?

    Thanx
  2. Subscriber coquette
    Already mated
    26 Mar '08 05:30
    i wish i was smart enough to know what is interesting about this
  3. Standard member adam warlock
    Baby Gauss
    26 Mar '08 10:49
    Originally posted by coquette
    i wish i was smart enough to know what is interesting about this


    Like a teacher of mine said: "The measure of interest in one thing is how much you try to understand it"

    In this case if the result holds it is a very interesting and general result. Of course it won't give us better toast in the morning but if you solve it you'll have the satisfaction of solving a mathematical theorem.
  4. 27 Mar '08 05:09
    this is basically sigma( n! / [(2n)!/2^n*n!]) or rather: sigma (2^n*n!*n!/[2n]!) as n goes from 1 to infinity, and then afterwards incrementing the denominator to get later series yes?

    i.e. more generally, looking at sigma (2^[n+k]*[n+k]!*n!/[2(n+k)]!) and looking for a convergence to some a*pi +b ?

    i haven't spent any time with pen and paper to see if this relates well to an earlier series you know with factoring, but i think the (2n+2k)! in the denominator grows faster than the numerator does, so we should be guaranteed convergence... can't be sure about a*pi +b with a,b rational though
  5. 05 Apr '08 12:01 / 2 edits
    I've managed to prove (in a fairly hand-wavy way) that these series converge to elements in Q[Pi] (the algebraic extension of the rationals by Pi). That doesn't get me any closer to whether that relation is linear, however... it seems plausible, but I haven't been able to prove that yet. Maybe when I get home I'll be able to use Mathematica to get something more for you. Thanks for the interesting problem!
  6. 05 Apr '08 12:51
    Originally posted by coquette
    i wish i was smart enough to know what is interesting about this
    We wish you were too!