Find ln(0.2) accurate to 0.001

Question

I want to find ln(0.2) accurate to 0.001. I get ln(1+(-0.8)) to solve the question. When I find the remainder I get the below one.

So I should find n for find ln(0.2). I find n=23 using a calculator by adding more values. I want to find it without adding more values. I want a mathematical method to find n.

Answer 1

$0.2 \mathrm{e}$ is closer to $1$ than is $0.2 \mathrm{e}^2$ and $\ln(0.2 \mathrm{e}) = \ln(0.2) + 1$. The partial sums in the alternating Taylor series for $\ln(1+x)$ bracket the value of $\ln(0.2 \mathrm{e})$ by less than $10^{-3}$ in $7$ terms.

If you permit yourself finer powers of $\mathrm{e}$ than just integer powers, you can accelerate convergence magnificently. For instance, applying the above to $\ln(0.2 \mathrm{e}^{\frac{37}{23}})$ yields the result in one term. Of course, one reasonably asks why you believe you can evaluate rational powers of $\mathrm{e}$ any more easily than evaluating these logarithms.

Answer 2

Finally I could get answer to Remainder. After that, How to find n?? https://i.stack.imgur.com/zdUep.jpg