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Math Help - Bounds of the Harmonic Series

  1. #1
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    Bounds of the Harmonic Series

    Could anyone suggest a neat way to proving that the Harmonic Series,
    Hn=1+1/2+1/3+1/4+...+1/n

    is bounded by ln(n){lower bound} and ln(n)+1{upper bound} ?

    ln is the natural logarithm.
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  2. #2
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    Quote Originally Posted by p vs np View Post
    Could anyone suggest a neat way to proving that the Harmonic Series,
    Hn=1+1/2+1/3+1/4+...+1/n

    is bounded by ln(n){lower bound} and ln(n)+1{upper bound} ?

    ln is the natural logarithm.
    Read this: http://faculty.prairiestate.edu/skifowit/htdocs/sd1.pdf
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  3. #3
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    Quote Originally Posted by p vs np View Post
    Could anyone suggest a neat way to proving that the Harmonic Series,
    Hn=1+1/2+1/3+1/4+...+1/n

    is bounded by ln(n){lower bound} and ln(n)+1{upper bound} ?

    ln is the natural logarithm.
    it's clear for n = 1. for n > 1 write mean value theorem for f(t)=\ln t, \ t > 1, on the interval [x-1,x] to get: \ln x - \ln (x-1)=\frac{1}{c}, for some x-1 < c < x. therefore:

    \frac{1}{x} < \ln x - \ln(x-1) < \frac{1}{x-1}. hence: \sum_{k=2}^n \frac{1}{k} < \sum_{k=2}^n(\ln k-\ln(k-1)) < \sum_{k=2}^n \frac{1}{k-1}, which gives us: H_n - 1 < \ln n < H_n -\frac{1}{n}. thus: \frac{1}{n}+\ln n < H_n < 1+\ln n, which

    is a better result than what you asked!
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  4. #4
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    Compact and elegant.
    Thanks
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  5. #5
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    here's another way of proving those bounds:

    for each k\in\mathbb N is \frac1{k+1}<\int_k^{k+1}\frac{dt}{t}<\frac1k.

    now it's \frac1n+\ln n=\frac1n+\int_1^n\frac{dt}{t}=\frac1n+\sum_{k=1}^  {n-1}\int_k^{k+1}\frac{dt}t<\frac1n+\sum_{k=1}^{n-1}\frac1k=\sum_{k=1}^n\frac1k.

    on the other hand is \sum_{k=1}^n\frac1k=1+\sum_{k=2}^n\frac1k=1+\sum_{  k=1}^{n-1}\frac1{k+1}<1+\sum_{k=1}^{n-1}\int_k^{k+1}\frac{dt}t=1+\int_1^n\frac{dt}t=1+\l  n n.

    finally it's \frac1n+\ln n<\sum_{k=1}^n\frac1k<1+\ln n.\quad\blacksquare
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