Being a little more careful than for the previous example, let's keep track of the error term in the example we've been doing: we have $${1\over 1-x}=1+x+x^2+\ldots+x^n+ {1\over (n+1)}{ 1 \over (1-c)^{n+1 }}x^{n+1}$$ for some $c$ between $0$ and $x$, and also depending upon $x$ and $n$. One way to avoid having the ${ 1 \over (1-c)^{n+1 }}$ ‘blow up’ on us, is to keep $x$ itself in the range $[0,1)$ so that $c$ is in the range $[0,x)$ which is inside $[0,1)$, keeping $c$ away from $1$. To do this we might demand that we integrate over the interval $[0,T]$ with $0\le T <1$.

For simplicity, and to illustrate the point, let's just take $0\le
T\le {1\over 2}$. Then in the *worst-case scenario*
$$\left|{ 1 \over (1-c)^{n+1 }}\right|\le { 1 \over (1-{1\over 2 })^{n+1}}=
2^{n+1}$$

Thus, *integrating the error term,* we have
\begin{align*}
\left|\int_0^T{1\over n+1}{ 1 \over (1-c)^{n+1 }}x^{n+1}\;dx\right|&\le
\int {1\over n+1}2^{n+1}x^{n+1}\;dx
={ 2^{n+1} \over n+1}\int_0^Tx^{n+1 }\;dx\\
&=\frac{ 2^{n+1}}{n+1}\left[{x^{n+2} \over n+2}\right]_0^T
={2^{n+1}T^{n+2} \over (n+1)(n+2) }
\end{align*}
Since we have cleverly required $0\le T\le {1\over 2}$, we actually
have
\begin{align*}
\left|\int_0^T{1\over n+1}{ 1 \over (1-c)^{n+1 }}x^{n+1}\;dx\right|&\le
{ 2^{n+1}T^{n+2} \over (n+1)(n+2) }\\
&\le
{ 2^{n+1}({1\over 2})^{n+2} \over (n+1)(n+2)}={1\over 2(n+1)(n+2) }
\end{align*}

That is, we have $$\left|-\log(1-T)-\left[T+{T^2\over 2}+\ldots+{T^n\over n}\right]\right|\le {1\over 2(n+1)(n+2)}$$ for all $T$ in the interval $[0,{1\over 2}]$. Actually, we had obtained $$\left|-\log(1-T)-\left[T+{T^2\over 2}+\ldots+{T^n\over n}\right]\right|\le { 2^{n+1}T^{n+2} \over 2(n+1)(n+2) }$$ and the latter expression shrinks rapidly as $T$ approaches $0$.