Prove that $|R_{n,0}| \leq \frac{|x|^{n+1}}{(n+1)(x+1)}$ 
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
$.
