How to prove $(n+1/2)\log (1+1/n)-1 >0$ [duplicate]

I want to show the following statement:

For any integer $n$ with $n\geq 1$, $$\left(n+\frac{1}{2}\right) \log \left(1+\frac{1}{n}\right) -1>0.$$ or equivalently, $$\log \left(n+1\right)-\log \left(n\right) >\frac{2}{2n+1 }. $$ Here log represents the natural log function. I am able to prove $$\log \left(n+1\right)-\log \left(n\right) >\frac{1}{n+1 }. $$ but not much luck with getting $\frac{2}{2n+1 }$. Any help is appreciated.

Solution 1:

For any integer $n$ with $n≥1$,$$\Big(n+\frac{1}{2}\Big)\log\Big(1+\frac{1}{n}\Big)-1>0$$ let $x=\frac{1}{n}$, then $$\Big(n+\frac{1}{2}\Big)\log\Big(1+\frac{1}{n}\Big)-1>0,n\geq1$$ $$\Leftrightarrow (\frac{1}{x}+\frac{1}{2})\log(1+x)-1>0,1\geq x>0$$ $$\Leftrightarrow \log(1+x)>\frac{2x}{x+2},1\geq x>0$$ let$$F(x)=\log(1+x)-\frac{2x}{x+2}$$ then:$$F'(x)=\frac{1}{x+1}-\frac{2(x+2)-2x}{(x+2)^2}=\frac{(x+2)^2-4(x+1)}{(x+1)(x+2)^2}=\frac{x^2}{(x+1)(x+2)^2}>0$$ so $F(x)$ is monotonically increasing function.So $F(x)>F(0)=0$.Done.