Instead of your series, it is rather usual to consider this one for computing values of $\ln$ function :
$$\ln \left(\dfrac{1+x}{1-x} \right)=2\left(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+\cdots \right)\tag{1}$$
with quicker convergence (see "more efficient series" in https://en.wikipedia.org/wiki/Logarithm).
With which value of $x$ ? You have to solve for that
$$\dfrac{1+x}{1-x}=0.97$$
giving $$x=-\tfrac{0.03}{1.97}\approx -0.015228426395939...$$
Now, consider the first terms of (1):
$$\begin{matrix}2x&=&-0.030456852791878\cdots&\\
2x^3/3&=&-0.000002354365183\cdots&\\
2x^5/5&=&-0.000000000327593\cdots&\\
2x^7/7&=& \; -0.000000000000054\cdots \ &etc.\\
&&\rule[5pt]{90pt}{1pt}\\
\ln(0.97)&=&-0.030459207484709\cdots&
\end{matrix}$$
What do we see ? That taking the first term ($2x$) in the series above with this value of $x$ gives a value with already 5 common decimal places with the target value $\ln(0.97)$.
The desired precision (at least 7 decimal places) is obtained by adding the two first terms.
Remarks :
1) Take a look at the many interesting answers to the following question What is the fastest algorithm for finding the natural logarithm of a big number?, in particular iterative Halley's method.
2) One finds on libraries' shelves dusty hand-computed "logarithm tables" ; it means strenuous efforts using different techniques (usually not involving series expansion (1)). An interesting book on these issues : "The History of Mathematical Tables. From Sumer to spreadsheets" (Oxford University Press) by Campbell-Kelly et al. It contains a wealth of interesting information in particular about the computation of these tables:(https://www.semanticscholar.org/paper/The-history-of-mathematical-tables-%3A-from-Sumer-to-Campbell-Kelly-Robson/ea87b0b23e9b551fc9793b548cee3102fd5754a2). See in particular its chapters 2 and 4.
3) In the same vein :
https://www.sciencedirect.com/science/article/pii/S0315086002000186. Besides telling the strange story of a father and his two daughters devoting 25 years of their life to the computation of tables that weren't published), this article contains an interesting example that I think valuable to reproduce. It will help to understand one of the techniques - among others - used to find logarithms of prime numbers.
Imagine you are looking for a good approximation of $\ln 1619$ (in fact, most tables in the past were dealing with base $10$ logarithms, but it doesn't matter, the explanation is common with the neperian logarithms). Assume that by chance (or by observation) it has been noticed that $1619$ is part of a decomposition ending with many nines (or ending with ''$000001$'' for example, which amounts to the same) ; here:
$$1155999999=3 \times 7 \times 11^2 \times 281 \times 1619$$
Rewrite it under the form :
$$\ln(1619)=\ln(A - 1) - \ln 3 - \ln 7 - 2 \ln 11 - \ln 281 \ \ \text{with} \ \ A=1156 \times 10^6$$
Now transform the first expression in order to be able to use the classical series :
$$\ln(A-1)=\ln(A)+\ln(1-\tfrac{1}{A})=\ln(A)+\left(\tfrac{1}{A}+\tfrac{1}{2A^2}+\cdots\right)$$
which here is strikingly convergent series (because $\tfrac{1}{A} \approx 10^{-9}$) yielding in this case a $10^{-18}$ accuracy by only using the two terms one finds in the last parenthesis above, being assumed that $\ln 3, \ \ln 5, \ \ln 11, \ \ln 281, \ \ln 10, \ \ln 1156$ have been previously computed with a similar accuracy.
4) Connected : my answer to a similar problem : https://math.stackexchange.com/q/3193225 using the alternated series theorem to guarantee a bound of the error (but unfortunately with low convergence).
