The objective is to prove that $\ln(1+x) = T(x)$ for $x \in (-1,1]$, and the way I thought of doing this is to show:
For $f(x) = \ln(1+x)$, that $\lim_{n\to \infty}R_n(x) = \lim_{n\to\infty}\int_a^{x}f^{(n+1)}(t)\cdot \frac{(t-x)^n}{n!}dt = 0 $ when $x \in (-1,1]$
Doing this and setting a = 0 as usual, I end up obtaining $\lim_{n \to \infty} \int_0^{x} \frac{(t-x)^n}{(t+1)^{n+1}}dt = 0$, which I cannot prove. How do I prove it, and if I did something wrong to get to this point, what is my mistake?
We know the general term of the Taylor series expansion of $\ln{x+1}=a_r=(-1)^{r+1} \frac{x^r}{r} $. Using the ratio test to test for convergence, ie when $$\lim _{r\to \infty}\lvert \frac{a_{r+1}}{a_r}\rvert<1$$ $$\lim _{r\to \infty}\lvert \frac{a_{r+1}}{a_r}\rvert=\lim _{r\to \infty}\lvert \frac{(-1)^{r+2} \frac{x^{r+1}}{r+1}}{(-1)^{r+1} \frac{x^r}{r}}\rvert=\lim _{r\to \infty}\lvert \frac{ {\frac{x^{r+1}}{r+1}}}{ {\frac{x^r}{r}}}\rvert=\lim _{r\to \infty}\lvert\frac{xr}{r+1}\rvert=\lim _{r\to \infty}\lvert\frac{x}{1+\frac{1}{r}}\rvert=\lvert x\rvert$$ So, the expansion converges if $-1<x<1$ as required. Hope that helped!