If we let \(S\) be the sum of the series \(\sum_{k=1}^{\infty}
\frac{(-1)^{k+1}}{k}\text{,}\) then we know that
\begin{equation*}
\left| S_{100} - S \right| \lt a_{101}\text{.}
\end{equation*}
Now
\begin{equation*}
a_{101} = \frac{1}{101} \approx 0.0099\text{,}
\end{equation*}
so the 100th partial sum is within 0.0099 of the sum of the series. We have discussed the fact (and will later verify) that
\begin{equation*}
S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln(2)\text{,}
\end{equation*}
and so \(S
\approx 0.693147\) while
\begin{equation*}
S_{100} = \sum_{k=1}^{100} \frac{(-1)^{k+1}}{k} \approx 0.6881721793
\end{equation*}
.
We see that the actual difference between \(S\) and \(S_{100}\) is approximately \(0.0049750013\text{,}\) which is indeed less than \(0.0099\text{.}\)