Integral remainder converges to 0
A related technique. Here is a reference for the remainder theorem. Since you are deriving the Taylor series at $0$, then $x_0=0$. Now,
$$ |R_{n+1}(x)|\leq \int_{0}^{x} (x-t)^n \frac{1}{(1+t)^{n+1}} dt \leq \int_{0}^{x} x^n \frac{1}{(1+t)^{n+1}} dt < \int_{0}^{1} \frac{1}{(1+t)^{n+1}} dt $$
$$ = \frac{1-2^{-n}}{n} \longrightarrow_{n\to \infty} 0 $$
Note that, we used the inequality $ x-t \leq x $ which follows from the fact
$$ 0\leq t \leq x \implies -x \leq -t \leq 0 \implies 0<x-t < x. $$