Why $\frac{1}{n+1}<\log(n+1)-\log(n)<\frac{1}{n}$?
logarithm inequality
4
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real-analysis
inequality
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3Because $e>(1+\frac{1}{n})^n$ and $e<(1+\frac{1}{n})^{n+1}$. Those are elementary inequalities: the first comes from the proof $(1+\frac{1}{n})^n$ is strictly increasing; the second from the fact that $(1+\frac{1}{n})^{n+1}$ is strictly decreasing; and both $(1+\frac{1}{n})^n,\ (1+\frac{1}{n})^{n+1}\to e$ as $n\to \infty$. – 2011-03-27
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0What conditions on n? (for instance, for n=-0.1 this is not true) – 2011-03-27
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0Just to clarify: my previous comment was addressed to the OP Vafa – 2011-03-27