I couldn't find a duplicate question, so I apologize if this has been asked before.
I'm trying to show that
$$ n – 1 < \left(\log \left( \frac{n}{n-1}\right)\right)^{-1} < n \tag{1} $$
I've verified this numerically, and it even seems to be the case that
$$ \lim_{n \to \infty} \frac{1}{\log \left( n / (n – 1)\right)} = n – \frac{1}{2} $$
Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then
$$ \log(n) – \log(n – 1) \sim \frac{1}{n} \tag{2} $$
However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.
Hints are definitely welcome.
Best Answer
Your inequality is equivalent to$$\frac1{n-1}>\log\left(\frac n{n-1}\right)>\frac1n,$$which, in turn, is equivalent to$$\frac1{n-1}>\log(n)-\log(n-1)>\frac1n.$$Now, use the fact that$$\log(n)-\log(n-1)=\int_{n-1}^n\frac{\mathrm dt}t.$$