Expert: Paul Klarreich - 4/27/2009

Question
The natural logarithm is defined by ln x = integral from 1 to x of 1/t dt.  By considering the subdivision 1, 2, ....,n of [1,n] and upper and lower approximations to the integral representing ln n based on this subdivision, show that

0 < integral from r=1 to n of (1/r) – ln n <= 1


Not sure how to start with this one.  I have no idea on what the upper and lower approximations would be???  Can you assist with this?

Thanks


Answer
Questioner:   Steve
Country:  Australia
Category:  Advanced Math
Private:  No
 
Subject:  Algebra - natural logs question
Question:  The natural logarithm is defined by ln x = integral from 1 to x of 1/t dt.  By considering the subdivision 1, 2, ....,n of [1,n] and upper and lower approximations to the integral representing ln n based on this subdivision, show that

0 < integral from r=1 to n of (1/r) – ln n <= 1


Not sure how to start with this one.  I have no idea on what the upper and lower approximations would be???  Can you assist with this?

Thanks
............................
Hi, Steve,
[I assume you have corrected your error regarding bounded sequences.]

Now upper and lower approximations (a.k.a. upper and lower sums -- see your elementary calculus book about Riemann sums) are:

UPPER SUM:  In your interval [k, k+1], choose the largest value of f(t) -- in the case of f(t) = 1/t, a decreasing function, take f(k).

LOWER SUM:  In your interval [k, k+1], choose the smallest value of f(t) -- in the case of f(t) = 1/t, a decreasing function, take f(k+1).

Your sum looks like this:

{n
|   (1/r) dr ~~=
}1

SUM(k=1 to n-1) f(r*) (1)    <-- the (1) is the width

and r* is your choice.  For:

UPPER:

SUM(k=1 to n-1) [1/k] = 1 + 1/2 + 1/3 + ... + 1/(n-1)


LOWER:

SUM(k=1 to n-1) [1/k] = 1/2 + 1/3 + ... + 1/(n-1) + 1/n


Now UPPER - LOWER = 1 - 1/n  and now that is clearly < 1 and > 0, provided  n > 1.  I think you can do something with that.