compute the lim as n approaches infinity (1/3n + 1/(3n + 1) + 1/(3n + 2) + ...+ 1/(4n -1))

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- Jun 3rd 2006, 10:46 AMSusie38Limit
compute the lim as n approaches infinity (1/3n + 1/(3n + 1) + 1/(3n + 2) + ...+ 1/(4n -1))

- Jun 3rd 2006, 11:13 AMCaptainBlackQuote:

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 ). - Jun 3rd 2006, 01:36 PMNichelle14
my professor said for a hint multiply each term by n/n. how does this help?