Let $r = \frac{12}{22}.$ It can be shown that
$$-\ln(1 - r) = \sum_{n=1}^{\infty} \frac1n r^n.$$
Let $$s_k = \sum_1^k \frac1n r^n.$$
(A.) Find the smallest number $M$ such that $s_k \leq M$ for every positive integer $k.$
$M =$
(B.) Find $s_3.$
$s_3 =$
(C.) Note that $1-r = \frac{10}{22}.$ Then $- \ln(1-r) = \ln(\frac {22}{10}).$
Suppose $s_3$ is used to approximate $\ln( \frac{22}{10}).$
The error is $\displaystyle \sum_{n=4}^{\infty} \left(\frac1n\right) r^n,$ which is less than $\displaystyle \left(\frac14\right)\sum_{n=4}^{\infty} r^n.$
Use the formula for the sum of a geometric series
to calculate this last sum and thereby to estimate the error in the approximation.
ERROR $\leq$
Your answer to C. should be more than the actual error which is 0.0401478186662987.