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?
 A: Exponentiate (formally, using the power series in x for exponentiation), subtract 1 and compare again. Knowing that the exp-function is entire and is strictly monotonic increasing over the reals suffices to make sure, that the greater/smaller-relation shall not be affected by that exponentiation.    
[add] The left comparision requires then the (given) property of x being "any positive integer" thus $\small x \ge 1$ to be also true (thanks to Didier's comment) [/add]
A: It's actually true for any real number $x\ge1$. Making the change of variables $t=1/x$, it's easier to show that $2\ln(1+t) > t > \ln(1+t)$ for any $0< t\le 1$.
The function $\ln(1+t)$ is (strictly) concave down (check its second derivative), and so every tangent line to its graph lies above its graph (strictly above the graph, except at the point itself). Since the tangent line at $t=0$ is simply $y=t$, that shows that $t > \ln(1+t)$ for any $t>0$.
On the other hand, $2\ln(1+t)$ is also concave down, and so the graph lies above any secant (any line segment connecting two points of the graph). Choosing the points $(0,0)$ and $(1,2\ln 2)$, we see that $2\ln(1+t) > (2\ln 2)t$ for all $0<t<1$. Since $2\ln2=\ln4>\ln e=1$, this gives $2\ln(1+t) > t$.
A: 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.
