Example 11.9.2 Starting with the geometric series: $$\eqalign{ {1\over 1-x} &= \sum_{n=0}^\infty x^n\cr \int{1\over 1-x}\,dx &= -\ln|1-x|+C = \sum_{n=0}^\infty {1\over n+1}x^{n+1}\cr \ln|1-x| &= C+\sum_{n=0}^\infty -{1\over n+1}x^{n+1}\cr }$$ We can figure out what $C$ is: When $x=0$, this becomes $\ln|1|=C+0$, so $C=0$, and $\ln|1-x| = \sum_{n=0}^\infty -{1\over n+1}x^{n+1}$, when $|x|< 1$. The series does not converge when $x=1$ but does converge when $x=-1$ or $1-x=2$. The interval of convergence is $[-1,1)$, or $0< 1-x\le2$, so we can use the series to represent $\ln(x)$ when $0< x\le2$. For example $$ \ln(3/2)=\ln(1- -1/2)= \sum_{n=0}^\infty (-1)^n{1\over n+1}{1\over 2^{n+1}} $$ and so $$ \ln(3/2)\approx {1\over 2}-{1\over 8}+{1\over 24}-{1\over 64} +{1\over 160}-{1\over 384}+{1\over 896} ={909\over 2240}\approx 0.406 .$$ Because this is an alternating series with decreasing terms, we know that the true value is between $909/2240$ and $909/2240-1/2048=29053/71680\approx .4053$, so correct to two decimal places the value is $0.41$.

What about $\ln(9/4)$? Since $9/4$ is larger than 2 we cannot use the series directly, but $$\ln(9/4)=\ln((3/2)^2)=2\ln(3/2)\approx 0.82,$$ so in fact we get a lot more from this one calculation than first meets the eye. To estimate the true value accurately we actually need to be a bit more careful. When we multiply by two we know that the true value is between $0.8106$ and $0.812$, so rounded to two decimal places the true value is $0.81$. $\square$