compute the lim as n approaches infinity (1/3n + 1/(3n + 1) + 1/(3n + 2) + ...+ 1/(4n -1))
Let:Originally Posted by Susie38
$\displaystyle
S_n=\frac{1}{3n}+\frac{1}{3n+1}+\dots+\frac{1}{4n-1}
$
Now (you will need to justify this, but its a standard trick):
$\displaystyle
\int_{3n}^{4n} \frac{1}{x-1}\ dx>S_n>\int_{3n}^{4n} \frac{1}{x}\ dx
$
Which will be sufficient to show that:
$\displaystyle \lim_{n \to \infty} S_n=\log(4/3)$
RonL
(A partial explanation of how the "trick" works can be found here ).