Prove: $\frac{x}{1+x} < \ln(1+x) < x$.

I thought a good practice would be to prove it using Taylor Expansion.

Here's my try:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3}...$$

The n=1 Taylor polynomial is: $$T_1(x) = x$$ and $$ ln(1+x) = T_1(x) + R_1(x)$$

Lets evaluate $R_1(x)$ by Cauchy's remainder formula:

$$R_1(x) = \frac{f^{(2)}(\xi)}{2!}\cdot x^2 = \frac{\frac{-1}{(\xi+1)^2}}{2!}\cdot x^2 = \frac{-x^2}{2(\xi+1)^2} < 0$$

Now, it does prove the right-hand side because $x + R_1(x) < x$ ($R_1(x)$ is negative). I'm not so sure what should I do for the left-hand side. I'd also like to get general critique for my current work.



We apply the mean value theorem on the function $t\mapsto \ln t$ on the interval $[1,1+x]$: there's $\zeta\in(1,1+x)$ such that $$\ln(1+x)=\frac x\zeta$$ and notice that $$\frac1{1+x}<\frac1\zeta<1$$

  • $\begingroup$ Very nice, thank you! $\endgroup$ – AnnieOK Jun 22 '14 at 16:00
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jun 22 '14 at 16:02

Using the elementary inequality $$ 1+x\le e^x $$ one directly obtains one side of the inequality chain. Replace $x$ by $-y$ and invert to obtain $$ \frac1{1-y}\ge e^y $$ and then set $1+x=\frac1{1-y}=1+\frac{y}{1-y}$ or $y=\frac{x}{1+x}$ to obtain $$ 1+x\ge e^{\frac{x}{1+x}} $$ for the other part of the inequality chain.



Assuming you argued and justified that $ f (x) = Ln (1 + x) $ is continuous and derivable $ \forall x> 0 $.

This is equivalent to writing the interval $ (0, x) $.

By the Mean Value theorem $\exists c \in (0,x)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$, i.e. $\frac{1}{1+c}=\frac{Ln(1+x)-Ln(1)}{x-0}=\frac{Ln(1+x)}{x}$.

We can delimit $\frac{1}{1+c}=f'(c)$ (*), then:


But we know that $f'(c)=\frac{1}{1+c}=\frac{Ln(1+x)}{x}$ (**)

Replacing in (*) we get:


Multiplying the inequality by $x$, we obtain the desired result:



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