How can I prove that $x-{x^2\over2}<\ln(1+x)$

Question

How can I prove that $$\displaystyle x-\frac {x^2} 2 < \ln(1+x)$$ for any $x>0$

I think it's somehow related to Taylor expansion of natural logarithm, when:

$$\displaystyle \ln(1+x)=\color{red}{x-\frac {x^2}2}+\frac {x^3}3 -\cdots$$

Can you please show me how? Thanks.

Answer 1

Hint:

Prove that $\ln(1 + x) - x + \dfrac{x^2}2$ is strictly increasing for $x > 0$.

edit: to see why this isn't a complete proof, consider $x^2 - 1$ for $x > 0$. It's strictly increasing; does that show that $x^2 > 1$? I hope not, because it's not true!

Answer 2

consider $f(x)=\ln(1+x)-x+\dfrac{x^2}{2}$

$f^{'}(x)=\dfrac{x^2}{1+x}>0$ forall $x>0$

Hence f(x)>f(0)