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?

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Both inequalities are immediate from the definition of $\ln$ as an area under a hyperbola, so answering this question really requires a starting point: what is your definition of $\ln$? –  whuber Oct 5 '11 at 5:08
See this previous question; doing the change of variable $t = 1+\frac{1}{x}$ yields one of your inequalities; the other one can be done similarly. –  Arturo Magidin Oct 5 '11 at 5:39

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]

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And use (crucially) the fact that 1/x is in (0,1], since otherwise the result is false. –  Did Oct 5 '11 at 5:07
ah, well... I've looked only on the right comparision seriously and missed to check the left comparision equivalently with attention, thanks! –  Gottfried Helms Oct 5 '11 at 5:33
OK, this works for the first inequality 1/x > ln(1+1/x). But for the second part... Working through gets me to trying to show e^(1/x) = 1 + 1/x + 1/2!x^2 + 1/3!x^3 + ... < (1 + 1/x)^2 , which isn't obviously true... How can I continue? –  DDTgg Oct 5 '11 at 5:53
@DDTgg: expand the quadratic, and subtract $\small 1+1/x+1/x^2/2!$ on both sides. Multiply with x. You'll get $\small 1/3!/x^2 + 1/4!/x^3 + ... < 1+1/x/2$ For $\small x \ge 1$ the rhs is "obviously" smaller than 1 because $\small 1/6 + 1/24 + 1/120 + \cdots$ is much smaller than $\small 1/2 + 1/4 +1/8+ \cdots$ –  Gottfried Helms Oct 5 '11 at 6:46

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 $2\ln(1+t) > t$.

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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.

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