This is the solution using the Lagrange remainder and its straightforward estimate. Tell me if this is not what you want.
The Taylor series centered at 𝑥 = 0 for the function 𝑓(𝑥) = ln(1 − 𝑥) is
𝑇(𝑥) = − ∑
The remainder of the Taylor 𝑛-th partial sum may be expressed as
𝑓(𝑥) − 𝑇𝑛 (𝑥) =
𝑓 (𝑛+1) (𝜉) 𝑛+1
(𝑛 + 1)!