Squeeze theorem inequality: $\frac{n}{1+n} < \ln(1+\frac{1}{n})^n < 1$

Question

show that for $n>1$, $$\frac{n}{1+n} < \ln(1+\frac{1}{n})^n < 1$$

by using $\frac{1}{n} < \ln n $ and $\ln(1+n) < n$ for $n>1$

I am unable to prove $\frac{n}{1+n} < \ln(1+\frac{1}{n})^n $

Answer 1

$$\frac{1}{\frac{n+1}{n}}=\frac{n}{n+1}<\ln(\frac{n+1}{n})=\ln(1+\frac{1}{n})<\frac{1}n$$

For $n>1$, we have $0<\ln(1+\frac{1}{n})<1, $ so $$\ln(1+\frac{1}{n})^n<1$$ Also, $$1>\ln(1+\frac{1}{n})^n=n*\ln(1+\frac{1}{n})>n(\frac{n}{n+1})>\frac{n}{n+1}\text{ , for }n>1$$ finishes the proof

Answer 2

One may observe that $$ f(x)=\ln(1+x)-\frac{x}{1+x},\qquad x \in[0,1], $$ is such that $$ f(0)=0,\quad f'(x)=\frac{x}{(1+x)^2}\ge0,\qquad x \in[0,1], $$ giving $$ \ln(1+x)-\frac{x}{1+x}\ge0,\qquad x \in[0,1]. $$ By putting $x:=\dfrac1n$, $n \ge1$, one gets $$ n\ln\left(1+\frac1n\right)-\frac{n}{1+n}\ge 0 $$ as wanted.