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....

@Michael Hardy offered a clear solution to the first part.
Second Part solution
The second part of the problem can be solved by applying the solution from the first part:
\begin{align*}
\left(1+\frac{1}{2n}\right)^{n+3} &< \left(1+\frac{1}{n}\right)^{n-1} \\
\ln \left(1+\frac{1}{2n}\right)^{n+3} &< \ln \left(1+\frac{1}{n}\right)^{n-1} &\text{($\ln$ is increasing function)} \\
(n+3) \ln \left(1+\frac{1}{2n}\right) &< (n-1) \ln \left(1+\frac{1}{n}\right) \\
(n+3) \left(\frac{1}{2n}-\frac{\left(\frac{1}{2n}\right)^2}{2}\right) &< (n-1) \left(\frac{1}{n}-\frac{\left(\frac{1}{n}\right)^2}{2}\right) \\
\frac{n+3}{2n}-\frac{n+3}{8n^2} &< \frac{n-1}{n}-\frac{n-1}{2n^2} \\
(4n^2+12n)-(n+3) &< (8n^2-8n)-(4n-4) \\
0 &< 4n^2-23n+7 \\
\end{align*}
(Algebra in $\LaTeX$. Yuck!)
We solve the quadratic for zeros (alternatively, plot it and eyeball the solution since we're estimating anyways) to get
$n \ge \frac{23 + \sqrt{417}}{8} \approx 5.428 \text{ or } n \le \frac{23 - \sqrt{417}}{8} \approx 0.322$. Since $n$ must be a positive integer, we solve $$n \ge 6$$
Additional Notes
Wolfram Alpha gives an "exact approximation" of $n \ge 5.357$ for the intersection point of the two functions. This value is, as we see, quite close to our polynomial-based approximation, because, based on a plot or a quick calculation, the second-degree approximation $\ln(1+x)\approx x-\frac{x^2}{2}$ is pretty darn close arount $x=0.2$.
If we try to solve this with the first-degree approximation $\ln(1+x) \approx x$, we get
\begin{align*}
(n+3)\left(\frac{1}{2n}\right) &< (n-1)\left(\frac{1}{n}\right) \\
\frac{5}{2n} &< \frac{1}{2} \\
5 &< n \\
n &\ge 6
\end{align*}
which, amusingly, is just enough accuracy for me to get the right answer. Should've done that, maybe, and saved myself from having to use the quadratic equation. A quick calculation shows that the first-degree approximation is not as good, but still somewhat good, as the second-degree approximation.
