Inequalities from Taylor expansions of $\log$ functions

The derivatives of $f(x) = \log (1+x)$ are $$ f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n} \quad (n \ge 1) $$ If we denote the $n$th Taylor polynomial with $T_n$ $$ T_n(x) = \sum_{k=1}^n \frac{(-1)^{n-1}}{n} x^n $$ and the remainder with $R_n$ $$ \log(1+x) = T_n(x) + R_n(x) $$ then Taylor's theorem with the mean-value form of the remainder gives $$ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1} = \frac{(-1)^nx^{n+1}}{(n+1)(1+\xi)^{n+1}} $$ for some $\xi$ between $0$ and $x$.

So in this case, $f^{(n)}(x)$ has alternating signs independent of $x$, and it follows that $$ \log(1+x) \begin{cases} < T_n(x) & \text{ for } -1 < x < 0 \\ < T_n(x) & \text{ for } 0 < x \le 1, n \text{ odd} \\ > T_n(x) & \text{ for } 0 < x \le 1, n \text{ even} \end{cases} $$

The case $-1 < x < 0$ is also obvious because all terms in the Taylor expansion are negative.

For $0 < x \le 1 $ it would also follow because the Taylor series is alternating with decreasing absolute values.

The same reasoning can be applied to $g(x) = (1+x)\log(1+x)$ because $g'(x) = 1 + \log(1+x)$, so that $g^{(n)}(x)$ has alternating signs for $n \ge 2$.

Alternating terms in the Taylor series alone are not sufficient to draw any conclusion about the relationship of $f(x)$ and $T_n(x)$, a simple counter example is $$ f(x) = x-\frac{x^2}2 + \frac{x^3}3 - x^4 \, , \quad T_2(x) = x-\frac{x^2}2 $$ with $$ f(x) \begin{cases} < T_2(x) & \text{ for }x < \frac 13 \\ > T_2(x) & \text{ for } x > \frac 13 \end{cases} $$