# Two-sided logarithm inequality

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.

• Mar 24 '19 at 20:06

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

• Awesome hint. Thanks! Mar 24 '19 at 20:25
• I'm glad I could help. Mar 24 '19 at 20:26

Let's try with a reductio ad absurdum :

1 disequality

Suppose that for some $$n$$:

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

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

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

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

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}

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