
Inequalities
QUESTION
$\displaystyle \forall x > 2$
Use the fact that $\displaystyle \frac{1}{x} < \log x  \log (x1) < \frac{1}{x1} $
to show that $\displaystyle s_n  1 < \log x < s_{n1} < s_n $
where $\displaystyle s_n = \sum^n_{n=1} \frac{1}{n}$
i.e. $\displaystyle s_n = 1 + .... + \frac{1}{n} $
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MY THOUGHTS:
Clearly $\displaystyle s_n  s_{n1} = \frac{1}{n} > 0 $ $\displaystyle \Rightarrow s_n > s_{n1} $
Also, $\displaystyle s_n  1 = \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} < \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n1} + 1 = s_{n1} $ $\displaystyle \Rightarrow s_n  1 < s_{n1} $
But I don't know how to show that log x fits in the middle of that inequality. Please help.

Have you noticed that $\displaystyle \sum\limits_{k = 2}^n {\left[ {\log (k)  \log (k  1)} \right]} = \log (n)$?
Think collapsing sums.

Not until now.
Excellent, thank you very much. Solved it. :)