Or search by topic
![]() |
Consider the area under the graph $y = 1/x$ between $x=a$ and $x=b$. This area lies between two rectangles and so we get $${b-a \over b} < \int_a^b{ 1\over x }dx = \ln b - \ln a < { b-a\over a}$$
If we evaluate the expression between $a = {1\over n}$ and $b= {1 \over {n-1}}$ we get:
|