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    By considering the area under the curve 1/x between k, k + 1, show that for k>0

    1/(k+1) < ln|(k+1)/k| < 1/k

    hence prove that

    ln(1 + N) < sigma {r = 1 to N} 1/r < 1 + lnN

    -----------


    First part was no problem, any direction on the latter? Please? It's 9am tomorrow and I just don't have this sorted yet. Pleaseee! Lex.
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    PLEASE. If there's no answer before I go tomorrow morning and one when I get back I will track you down and stab you in the face.
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    Well, I can get the left hand side of the inequality. The first part gives us

    1/(r+1) < ln(r+1) - ln(r) < 1/r

    Summing from r=1 to N

    SUM{r=1 to N} [ ln(r+1) - ln(r) ] < SUM{r=1 to N} 1/r

    The LHS is

    ln2 - ln1 +
    ln3 - ln2 +
    ln4 - ln3 +
    ... +
    ln(N-1) - ln(N-2) +
    ln(N) - ln(N-1) +
    ln(N+1) - ln(N)

    and cancelling terms gives ln(N+1)-ln1 = ln(N+1). So,

    ln(1+N) < SUM{r=1 to N} 1/r

    I don't yet see how to get the RHS of the inequality.
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    I don't know if this is the 'proof' they are looking for but the RHS of the inequality will follow from considering areas once again. We have

    SUM{r=1 to N} 1/r = 1 + 1/2 + 1/3 + 1/4 + ... + 1/N

    Now, let us interpret this as a sequence of rectangles each with a width of 1 and a height 1/r. If we arrange these such that they decrease in height from left to right and form what looks like one long 'block of stairs' we should see that the resulting shape when placed on the x-axis touching the y-axis and reaching up to the point x=N will fit all underneath the graph of y = 1/x. Now, note that ln(N) is the area under the graph y = 1/x from the line x=1 to the line x=N. Therefore, 1+ln(N) is the area bounded by the curve y=1/x, the x-axis, y-axis, the line x=N and the line y=1 (I think you are going to have to draw this yourself). So, it is now easily seen that the area of the stair shape is less than the area of the region just described. Therefore, the sum is less than 1+ln(N) and we have established the inequality

    ln(1+N) < SUM{r=1 to N} 1/r < 1 + ln(N)

    I hope I explained myself well. This constitutes a proof/justification; however, I am fairly sure this is not what they intended.
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    (Original post by mikesgt2)
    I don't know if this is the 'proof' they are looking for but the RHS of the inequality will follow from considering areas once again. We have

    SUM{r=1 to N} 1/r = 1 + 1/2 + 1/3 + 1/4 + ... + 1/N

    Now, let us interpret this as a sequence of rectangles each with a width of 1 and a height 1/r. If we arrange these such that they decrease in height from left to right and form what looks like one long 'block of stairs' we should see that the resulting shape when placed on the x-axis touching the y-axis and reaching up to the point x=N will fit all underneath the graph of y = 1/x. Now, note that ln(N) is the area under the graph y = 1/x from the line x=1 to the line x=N. Therefore, 1+ln(N) is the area bounded by the curve y=1/x, the x-axis, y-axis, the line x=N and the line y=1 (I think you are going to have to draw this yourself). So, it is now easily seen that the area of the stair shape is less than the area of the region just described. Therefore, the sum is less than 1+ln(N) and we have established the inequality

    ln(1+N) < SUM{r=1 to N} 1/r < 1 + ln(N)

    I hope I explained myself well. This constitutes a proof/justification; however, I am fairly sure this is not what they intended.
    No that seems to be 100% spot on, entirely in the usual spirit of these type of questions. Excellent, straightforward proof, here's hoping I can apply it if something similar comes up in a couple of hours.

    THanks so much.

    Alex
 
 
 
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