<s>[INST] Show $\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ I'm reading a proof which aim to show that:
$$\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$$
the Taylor expansion of $\ln(1+x)$ is (not by chance):
$$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}...$$
Now, the proof claims that the remainder, starting from $\frac{x^5}{5}$ is positive.
But, calculating the remainder using the Lagrange's form gives:
$$\frac{\ln(1+x)^{(6)}(z)}{6!}x^6 < 0$$  
Because $$\ln(1+x)^{(6)} = -\frac{120}{(x+1)^6} < 0$$
Where is the mistake?
 [/INST] There exists $c\in ]0,x[$ such that $\ln(1+x)=\ln(1+0)+\ldots+-\frac{x^4}{4}+\ln(1+c)^{(5)}\frac{x^5}{5!}$
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