By differentiating $\log x$ and noticing the pattern, we see that for all $n\geq 0$, $f^{(n+1)}(x)=(-1)^{n}n!x^{-(n+1)}$.

From the mean-value form of the remainder from Taylor's theorem, we know that there exists some $c$ between 1 and 1.1 such that

${\begin{aligned}f(1.1)-T_{n}(1.1)&={\frac {f^{(n+1)}(c)}{(n+1)!}}(1.1-1)^{n+1}\\&={\frac {(-1)^{n}}{(n+1)c^{n+1}}}\left(0.1\right)^{n+1}\\&={\frac {-1}{(n+1)c^{n+1}}}(-0.1)^{n+1}\end{aligned}}$

If $T_{n}(1.1)$ is an underestimate, then $T_{n}(1.1)<f(1.1)$, so $f(1.1)-T_{n}(1.1)>0$;
since $c>0$, the condition $f(1.1)-T_{n}(1.1)>0$ is equivalent with

${\begin{aligned}0<{\frac {-1}{(n+1)c^{n+1}}}(-0.1)^{n+1}\Longleftrightarrow 0>{\frac {1}{(n+1)c^{n+1}}}(-0.1)^{n+1}\Longleftrightarrow 0>(-0.1)^{n+1}.\end{aligned}}$

This occurs precisely when $n+1$ is odd; that is, when $n$ is even.

So, when $n$ is even, $T_{n}(1.1)$ is an underestimate for $f(1.1)$; in other words, when $\color {blue}n\in \{2m:m\in \mathbb {N} \}=\{0,2,4,6,8,\dots \}.$