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Math Help - cal problem #2

  1. #1
    Junior Member
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    cal problem #2

    if n is greater than or equal to 2, show that

    1/2 + 1/3 + ... + 1/n is less than or equal to ln n is less than or equal to 1 + 1/2 + 1/3 + ... + 1/ n-1
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    if n is greater than or equal to 2, show that

    1/2 + 1/3 + ... + 1/n is less than or equal to ln n is less than or equal to 1 + 1/2 + 1/3 + ... + 1/ n-1
    to prove:
    1/2 + 1/3 + ... + 1/n <= ln n <= 1 + 1/2 + 1/3 + ... + 1/n-1
    let 1/2 + 1/3 + ... + 1/n= a
    ln n =b
    1+ 1/2 + 1/3 + ... + 1/n-1=c
    all a,b,c are positive
    hence the question is equivalent to e^a <=e^b <= e^c
    proof of e^b<= e^c
    e^b =n
    we know that, e^x= 1 + x+ some positive quantity
    hence, e^x > 1+x (x is a positive number)
    now, e^c =e^(1 + 1/2 + 1/3 + ... + 1/n-1)
    =e^1*e^(1/2)...*e^(1/(n-1))
    >(1+1)(1+1/2)....(1+1/(n-1))
    >2*3/2*4/3.........*n/(n-1)
    >n=e^b
    proof of e^a<= e^b
    for x<1 e^x < 1+x+x^2 + x^3 +..........
    hence e^x < 1/(1-x)
    e^a= e^(1/2)e^(1/3)e^(1/4).....e^(1/n)
    <1/(1-1/2)*1/(1-1/3)*1/(1-1/4)*.......*1/(1-1/n)
    < 2*3/2*4/3*.....*n/(n-1)
    <n=e^b
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