Transforming the inequality bellow

Question

How do I get from here:

$$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^n+1$$

To this this:

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

Answer 1

First part: We have $$\left(1+\frac{1}{n}\right)^n<e$$ Taking the natural logarithm on both sides we get $$n\ln\left(\frac{n+1}{n}\right)<1$$ so $$n\left(\ln(n+1)-\ln(n)\right)<1$$ so $$\ln(n+1)-\ln(n)<\frac{1}{n}$$ for $$n>0$$ For the second part: We have $$e^{n/(n+1)}<\frac{n+1}{n}$$ so $$\frac{n}{n+1}<\frac{1}{e^{(n+1)/n}}=e^{-n/(n+1)}$$ so $$\ln(n)-\ln(n+1)<-\frac{n}{n+1}$$ or $$\ln(n+1)-\ln(n)>\frac{n}{n+1}$$