How do I show that $2\ln(1+1/x) > 1/x > \ln(1+1/x)$ for any positive integer $x$? How do I show that $$2\ln\left(1+\frac1x\right) > \frac1x > \ln\left(1+\frac1x\right)$$ for any positive integer $x$?
I know it's true but how do I show it?

Let's observe:  $\ln(1+\frac{1}{x})<\frac{1}{x}$
We can define $y=\ln(1+\frac{1}{x})-\frac{1}{x}$ ,Now we may find first derivative $y'$:
$y'=\frac{1}{1+\frac{1}{x}}(1+\frac{1}{x})'-(\frac{1}{x})' \Rightarrow y'=\frac{1}{x^2}(\frac{1}{x+1}) \Rightarrow y'>0$ for all positive integers..so function $y$ increases as $x$ increases for all $x$
Next,we will find horizontal asymptote as $\lim_{x \to \infty} y$
$\lim_{x \to \infty} ( \ln(1+\frac{1}{x})-\frac{1}{x})=\ln(1)=0$ so $y=0$ is horizontal asymptote.
Since $y$ increases and has $y=0$ horizontal asymptote it must be that $y<0$ for all $x$ which means 
$\ln(1+\frac{1}{x})<\frac{1}{x}$ is true for all positive $x$
For inequality $\frac{1}{x}<2\ln(1+\frac{1}{x})$ we may use next proven inequality which states that
$\frac{n}{n+1}<\ln(n+1)$  ,In your case this means that $2\ln(1+\frac{1}{x})>\frac{\frac{2}{x}}{\frac{1}{x}+1}=\frac{2}{x+1}$ so we have to show :
$\frac{1}{x}<\frac{2}{x+1} \Rightarrow \frac{x-1}{x(x+1)}>0$ which is true for all $x>1$, for $x=1$ we have that $\ln(2)>\frac{1}{2}$ This means that inequality:
$\frac{1}{x}<2\ln(1+\frac{1}{x})$ is true for all positive integers.
