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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    02 Apr '09 18:46
    Im trying to show that the series

    sum(n=1_infinity) 1/(4n^(1/3) -1) > sum(n=1_infinity) 1/n term by term

    As part of the proof of divergence by direct comparison

    so I need to show that

    n > 4n^(1/3) -1 in order for 1/n < 1/(4n^(1/3) -1)

    so i will compare the rates of change of the denominators ( both series are decreasing on the interval [1,inf.))

    n > 4n^(1/3) -1

    by taking the first derivative of both sides

    1 > 4/(3*n^(2/3))

    take the limit of both sides as n--> infinity

    1 > 0

    so there exists some value N[1, inf.) that, n >/= 4n^(1/3) -1

    which would show that for value greater than N[1,inf) 1/n </= 1/ (4n^(1/3) -1)

    or so I think....
  2. Standard member adam warlock
    Baby Gauss
    02 Apr '09 23:23 / 8 edits
    Originally posted by joe shmo
    Im trying to show that the series

    sum(n=1_infinity) 1/(4n^(1/3) -1) > sum(n=1_infinity) 1/n term by term

    As part of the proof of divergence by direct comparison

    so I need to show that

    n > 4n^(1/3) -1 in order for 1/n < 1/(4n^(1/3) -1)

    so i will compare the rates of change of the denominators ( both series are decreasing on the interval [1,i ...[text shortened]... would show that for value greater than N[1,inf) 1/n </= 1/ (4n^(1/3) -1)

    or so I think....
    If you rearrange the terms you get that what you want to prove is that (n+1)/4 > n^(1/3) which is equivalent to
    [; \left( \frac{n+1}{4}\right)^3 > n \Leftrightarrow \frac{(n+1)^3}{64} > n \Leftrightarrow \frac{n^3+3n^2+3n+1}{64} > n \Leftrightarrow \frac{n^3+3n^2-61n}{64} > 0 \Leftrightarrow ;]

    [; \Leftrightarrow n^3+3n^2-61n > 0 \Leftrightarrow n(n^2+3n-61) > 0 \Leftrightarrow n^2+3n-61 > 0 ;]

    But I think that this quadratic has real zeros and so either what you are trying to prove is wrong or I made a mistake...

    Use this in order to see my formulas: http://thewe.net/tex/
  3. Subscriber joe shmo On Vacation
    Strange Egg
    03 Apr '09 00:16
    Originally posted by adam warlock
    If you rearrange the terms you get that what you want to prove is that (n+1)/4 > n^(1/3) which is equivalent to
    [; \left( \frac{n+1}{4}\right)^3 > n \Leftrightarrow \frac{(n+1)^3}{64} > n \Leftrightarrow \frac{n^3+3n^2+3n+1}{64} > n \Leftrightarrow \frac{n^3+3n^2-61n}{64} > 0 \Leftrightarrow ;]

    [; \Leftrightarrow n^3+3n^2-61n > 0 \Leftrightarrow ...[text shortened]... made a mistake...

    Use this in order to see my formulas: http://thewe.net/tex/
    hmmm...well I didn't download the software

    but the polynomial n^3 +3n^2 -61n +1 = 0

    has 3 real solutions

    And there is only one in the domain that i am worried about ( it occurs between n= 6 & n=7 if i remember correctly

    so which ever series is above will continue to be above for the rest of the domain in question....or at least I hope..
  4. 03 Apr '09 07:39
    Okay, returning to your original idea, it is quite simple to show that for n>8 it is fulfilled n>4n^1/3 -1, or n-4n^3+1>0. Let's substitute n^1/3 = x. Then we have x^3 -4x>-1. Since x is positive and greater or equal to one (n is greater than 1) we have (x-2)(x+2)x>-1. The LHS is always positive for x>2 (n>8).
  5. Standard member Palynka
    Upward Spiral
    03 Apr '09 09:58
    First show that:
    1/(4n^(1/3)-1) > 1/n, for all n>6

    Then highlight the fact that H_inf - H_6 = inf (note H_n:harmonic number)

    So sum(n=7_infinity) 1/n < sum(n=7_infinity) 1/(4n^(1/3) -1) = inf
    => sum(n=1_infinity) 1/(4n^(1/3) -1) = inf
  6. 05 Apr '09 03:33 / 2 edits
    [for notation purposes i will use G(x) to replace "sum of all x as x begins at 1 and increases by one, tending towards infinity" i.e. it is the normal capital sigma we use with n=1 below and infinity on top.]

    you are looking to show that G(1/[4n^(1/3)-1]) diverges. so we wish to compare it term by term to G(1/n).

    first, however, consider G(1/4n)... if you write it out you will see quickly that G(1/4n) = 1/4 * G(1/n), which clearly diverges.

    but then term by term comparison is much easier... for all n greater than or equal to 1, 4n^(1/3) < 4n. furthermore, [4n^(1/3) - 1] < 4n^(1/3) < 4n.

    so then, every denominator in our series is strictly smaller than that of a divergent series, and every numerator is the same as that divergent series. so our series is term-by-term larger than a divergent series, and so also must diverge.

    note that in dealing with comparisons to the harmonic series, one of the most useful things to note is that a constant multiplier that does not increase as n increases is inconsequential to the divergence or convergence of the sequence. and so for a term-by-term comparison test it's often useful to multiply the harmonic series by a useful constant of our choice.

    cheers!
  7. Subscriber joe shmo On Vacation
    Strange Egg
    05 Apr '09 06:53
    Originally posted by Aetherael
    [for notation purposes i will use [b]G(x) to replace "sum of all x as x begins at 1 and increases by one, tending towards infinity" i.e. it is the normal capital sigma we use with n=1 below and infinity on top.]

    you are looking to show that G(1/[4n^(1/3)-1]) diverges. so we wish to compare it term by term to G(1/n).

    ...[text shortened]... s often useful to multiply the harmonic series by a useful constant of our choice.

    cheers![/b]
    ok..will ponder in light of information...but for now cheers as well!!!