The problem is: Prove that if $-1 < x \leq 0$, the the Taylor's remainder $R_{n,0}$ corresponding to $f(x) = \log(x+1)$ satisfies $$\displaystyle |R_{n,0}| \leq \frac{|x|^{n+1}}{(n+1)(x+1)} $$
First, I know that Taylor's remainder formula is $$\displaystyle R_{n,a} (x) = \frac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}$$ Knowing that $$ f^{(n+1)}(x) = \frac{(-1)^n n!}{(x+1)^{n+1}}$$ So, for $\log(x+1)$, $$ R_{n,0}(x) = \frac{(-1)^n x^{n+1}}{(n+1)(s+1)^{n+1}} $$ Then, $$ |R_{n,0}|= |\frac{(-1)^n x^{n+1}}{(n+1)(s+1)^{n+1}}| \leq \frac{|(-1)^n| |x^{n+1}|}{|(n+1)||(s+1)^{n+1}|} = \frac{|x^{n+1}|}{|(n+1)||(s+1)^{n+1}|}$$ I got stuck here, because I don't see where to go from this. Even if I rename $s$ as $x$, I see that $\displaystyle \frac{|x|^{n+1}}{(n+1)(x+1)}$ does not limit $\displaystyle \frac{|x|^{x+1}}{(n+1)(x+1)^{n+1}}$.
Any suggestion/help?
Here is an easy way to get the series for $\ln(1-t)$.
$\sum_{k=0}^{n-1} t^k =\dfrac{1-t^n}{1-t} $ so $\dfrac1{1-t} =\sum_{k=0}^{n-1}t^k+\dfrac{t^n}{1+t} $.
Integrating from $0$ to $x$,
$\begin{array}\\ \int_0^x\dfrac1{1-t}dt &=\sum_{k=0}^{n-1}\int_0^x t^kdx+\int_0^x \dfrac{t^n}{1-t}dt\\ \text{or}\\ -\ln(1-x) &=\sum_{k=0}^{n-1}\dfrac{x^{k+1}}{k+1}+\int_0^x \dfrac{t^n}{1-t}dt\\ \end{array} $
so
$\begin{array}\\ -\ln(1-x) -\sum_{k=0}^{n-1}\dfrac{x^{k+1}}{k+1} &=\int_0^x \dfrac{t^n}{1-t}dt\\ &=E_n(x)\\ \end{array} $
$\begin{array}\\ E_n(x) &=\int_0^x \dfrac{t^n}{1-t}dt\\ &\gt \int_0^x t^ndt\\ &=\dfrac{x^{n+1}}{n+1}\\ E_n(x) &=\int_0^x \dfrac{t^n}{1-t}dt\\ &\lt \int_0^x \dfrac{t^n}{1-x}dt\\ &=\dfrac{x^{n+1}}{(n+1)(1-x)}\\ \end{array} $
Put $n=1$. Then $E_1(x) =-\ln(1-x)-x $ and $\dfrac{x^2}{2} \lt E_1(x) \lt \dfrac{x^2}{2(1-x)} $.
If $0 < x \le \frac12$, then $E_1(x) \lt x^2 $.