Show $\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$

Question

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?

Answer 1

You can use MVT to do. In fact, let $$ f(x)=\ln(1+x)-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}). $$ Then, by MVT, there is $c\in(0,x)$ such that $$ f(x)-f(0)=f'(c)x=\frac{c^5}{1+c}x>0 $$ So $f(x)>f(0)$ for $x>0$ or $$\ln(1+x)>x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}. $$

Answer 2

There exists $c\in ]0,x[$ such that $\ln(1+x)=\ln(1+0)+\ldots+-\frac{x^4}{4}+\ln(1+c)^{(5)}\frac{x^5}{5!}$

Answer 3

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 $x > 0$.