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Interesting sums

  1. May 4, 2003 #1
    interesting inequality involving sums

    Which is the smallest n-natural number- for this inecuation:

    &#8721k=2n {1/[k * ln(k)]} &#8805 20

    Any ideas?
     
    Last edited: May 4, 2003
  2. jcsd
  3. May 4, 2003 #2
    1.0488269074484
    Hm... I wonder how i got that.
     
  4. May 4, 2003 #3
    Very interesting indeed...but how did you get that?'n' must be an integer...
    In case it is not clear the sum is from k=2 to n.
     
    Last edited: May 4, 2003
  5. May 4, 2003 #4
    I do not see where that number comes into play. It is certainly not the answer.
     
  6. May 5, 2003 #5
    I crawl back into my hole and will read the question properly next time.
     
  7. May 8, 2003 #6
    Re: interesting inequality involving sums

    Nonexistent.
    I think if we let n go to infinity we'll still never get above 5 for the sum.
     
  8. May 8, 2003 #7
    very easy
    u just use the integral test
    because the sum is continuous and decreasing
    if n approach to infinite
    the sum is diverges and must greater than 20
     
  9. May 8, 2003 #8
    I believe newton1 is correct in assuming the series diverges by integral test:

    Let f(x) = 1/( x ln(x) ), then since f(x) is positive, continous, and decreasing for n>=2 we apply Integral test.

    Integral[ 2->inf, 1/(x lnx) ] diverges, therefore the series diverges as well.

    Since the partial sums in this series are non-decreasing (all the terms in this series are positive) , then there will exist a least natural number N for which the sum will exceed 20 (and then stay above it).

    I'm not sure what the value of N is, but by applying a variant on the integral test, I think i've found a real value B for which N must be greater.

    Suppose you approximate the following integral with right-hand riemann sums with delta_x = 1:
    Integral[ 2->B, 1/(x lnx) ]
    Then the riemann sum approximation will be the sum from k=3 to B of 1/ ( n ln(n) ) (assuming B is integer) and this approximation will be an underestimate.

    From which it follows:
    Sum[k=2->B, 1/ (n ln(n) )] <
    Integral[2->B, 1/ (x lnx )] + 1/(2ln2)

    The value of N must be greater than the value of B that first causes the integral + 1/(2ln2) to go over 20 because the sum will be less than the integral + 1/(2ln2).

    Integral[ 2->B, 1/(x lnx) ] + 1/(2ln2) >= 20

    ln(ln(B)) - ln(ln(2)) +1/(2ln2) >= 20
    B >= e^e^(20- 1/(2ln2) +ln(ln(2)))
    B >= 2.726413 * 10^70994084

    This value is beyond astronomical!
    And if I've done my calculations correctly, N must be greater than this!
     
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