Inequalities

• Nov 12th 2008, 09:20 AM
WWTL@WHL
Inequalities
QUESTION

$\forall x > 2$

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

to show that $s_n - 1 < \log x < s_{n-1} < s_n$

where $s_n = \sum^n_{n=1} \frac{1}{n}$

i.e. $s_n = 1 + .... + \frac{1}{n}$

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MY THOUGHTS:

Clearly $s_n - s_{n-1} = \frac{1}{n} > 0$ $\Rightarrow s_n > s_{n-1}$

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

But I don't know how to show that log x fits in the middle of that inequality. Please help.
• Nov 12th 2008, 09:29 AM
Plato
Have you noticed that $\sum\limits_{k = 2}^n {\left[ {\log (k) - \log (k - 1)} \right]} = \log (n)$?
Think collapsing sums.
• Nov 12th 2008, 09:46 AM
WWTL@WHL
Not until now.

Excellent, thank you very much. Solved it. :)