If $\ln(1+x) \approx A+Bx+Cx^2$, differentiate twice both sides and show that $\ln(1+x) \approx x-\frac{1}{2}x^2$ 

Question: $\ln(1+x) \approx A+Bx+Cx^2$, for $-1<x\leq1$, where $A,B,C$ are constants.
Differentiate twice both sides of the approximation above and hence show that $$ \ln(1+x) \approx x-\frac{1}{2}x^2$$
Using logs, estimate the smallest value of the positive integer $n$ for which  $$ \left(1+\frac{1}{2n} \right)^{n+3} < \left(1+\frac{1}{n} \right)^{n-1}$$


My working:
For the first part: 
$$\ln(1+x) \approx A+Bx+Cx^2$$
$$ \Leftrightarrow \frac{1}{1+x} \approx B+2Cx$$ 
$$ \Leftrightarrow  \frac{-1}{(1+x)^2} \approx 2C $$
But now I do not know what to do with this....

\begin{align}
\ln(1+x) & = A + Bx +Cx^2. & & \text{When } x = 0 \text{ then } \ln 1 = A. \\
& & & \text{So } A=\ln 1 = 0. \\[10pt]
\frac 1 {1+x} & = B + 2Cx. & & \text{When } x=0 \text{ then } \frac 1 {1+0} = B. \\
& & & \text{So } B = 1. \\[10pt]
\frac {-1}{(x+1)^2} & = 2C. & & \text{When } x=0 \text{ then } \frac{-1}{(1+0)^2} = 2C. \\
& & & \text{So } C= \frac{-1}2.
\end{align}
Thus
$$
\ln(1+x) = x - \frac{x^2} 2 + \text{higher-degree terms}.
$$
After finding the derivatives on both sides you need to set $x=0$ on both sides, at each step finding one coefficient.
The higher-degree terms are microscopic compared to the initial terms if $x$ is very close to $0$.
