QUESTION

$\displaystyle \forall x > 2$

Use the fact that $\displaystyle \frac{1}{x} < \log x - \log (x-1) < \frac{1}{x-1} $

to show that $\displaystyle s_n - 1 < \log x < s_{n-1} < 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_{n-1} = \frac{1}{n} > 0 $ $\displaystyle \Rightarrow s_n > s_{n-1} $

Also, $\displaystyle s_n - 1 = \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} < \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n-1} + 1 = s_{n-1} $ $\displaystyle \Rightarrow s_n - 1 < s_{n-1} $

But I don't know how to show that log x fits in the middle of that inequality. Please help.