Two-sided logarithm inequality

Question

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.

Answer 1

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$$\log(n)-\log(n-1)=\int_{n-1}^n\frac{\mathrm dt}t.$$

Answer 2

Let's try with a reductio ad absurdum :

1 disequality

Suppose that for some $n$:

$$\log^{-1}(\frac{n}{n-1})<n-1 $$

Notice that $\log^{-1}(\frac{n}{n-1})=\log_{\frac{n}{n-1}}(e)$ so:

$\log_{\frac{n}{n-1}}(e)<n-1$

$(\frac{n}{n-1})^{n-1}>e$

Now $ n-1=x $:

$(1+\frac{1}{x})^{x}>e$

But this is absurd because $(1+\frac{1}{x})^{x}$ is strictly increasing and his limit is $e$.

2 disequality As before: $$\log^{-1}(\frac{n}{n-1})>n $$

$\log_{\frac{n}{n-1}}(e)>n$

$(\frac{n}{n-1})^{n}<e$

And this is absurd because that function is strictly decreasing and his limit value is $e$