Showing that the integral remainder of the Taylor expansion of $f(x)=-\log(1-x)$ goes to $0$

Question

Let $|x|<1$. Define $R_n(x):=\int_{0}^{x}\frac{(x-t)^{n-1}}{(1-t)^n}dt$. How do we prove that $\lim_{n\to \infty}R_n(x)=0$? This is actually the integral remainder of the Taylor expansion of the function $f(x)=-log(1-x)$. Once I show that $R_{n-1}(x)\to 0$, then I can say that $f(x)=\sum_{k=1}^{\infty}\frac{x^k}{k}$.

By the way, I am using the following formula: $$f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!} x^k+ R_{n-1}(x)$$ where $R_{n-1}(x)=\frac{1}{(n-1)!}\int_{0}^{x}(x-t)^{n-1} f^{(n)}(t)dt$

Answer 1

For $x > 0$ and $0 \leqslant t \leqslant x$, we have

$$\frac{x-t}{1-t} = 1 - \frac{1-x}{1-t} \leqslant 1 - \frac{1-x}{1} = x,$$

and for $x < 0$ and $x \leqslant t \leqslant 0$, we have

$$\left\lvert \frac{x-t}{1-t}\right\rvert = \frac{\lvert x\rvert - \lvert t\rvert}{1+\lvert t\rvert} \leqslant \frac{\lvert x\rvert}{1},$$

so in both cases, the integrand is bounded by

$$\frac{\lvert x\rvert^{n-1}}{1-\lvert x\rvert}$$

and for the integral we obtain the bound

$$\frac{\lvert x\rvert^n}{1-\lvert x\rvert}$$

which is known to tend to $0$ for $n\to \infty$ and $\lvert x\rvert < 1$.

Answer 2

As $$ r=\frac{|x-t|}{|1-t|}<1,\\ \frac 1{1-t}r^{n-1}\to 0 $$ and the function inside the integral has point wise limit $0$.

As it is bounded, you can apply the dominated convergence theorem, and directly get that the integral has limit $0$ as well.