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?

A much simpler proof is as follows. Let $t > 0$ and then we have $1 - t^{4} < 1$ so that on division by $(1 + t)$ we get $$(1 - t)(1 + t^{2})< \frac{1}{1 + t}$$ or $$1 - t + t^{2} - t^{3} < \frac{1}{1 + t}$$ If $x > 0$ then we can integrate the above inequality to get $$\int_{0}^{x} \left(1 - t + t^{2} - t^{3}\right)\,dt < \int_{0}^{x}\frac{dt}{1 + t}$$ and we get $$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} < \log(1 + x)$$ for $x > 0$.
