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....
 A: Other answers have offered the solution as expected. I wanted to highlight that such sort of questions tend to encourage more hand waving than doing any service to mathematics.
When you write $$\log (1 + x) \approx A + Bx + Cx^{2}\tag{1}$$ you actually mean that there is an error $R(x)$ involved so that $$\log(1 + x) = A + Bx + Cx^{2} + R(x)\tag{2}$$ for all values of $x$ with $|x| < 1$. It is not possible to differentiate both sides of a relation like $(1)$ which holds only approximately. Rather we can only differentiate both sides of identities like $(2)$ and then the further analysis depends on the nature and behavior of error term $R(x)$ and its derivatives. The differentiation of approximation cannot be justified because we can't ensure that if $R(x)$ is small then its derivative will also be small for the values of $x$ under consideration.
A proper way to obtain the approximation for $\log(1 + x)$ is to apply Taylor's theorem or perhaps directly start with the approximation $$\frac{1}{1 + t} \approx 1 - t$$ and then see this approximation as an identity with error term as $$\frac{1}{1 + t} = 1 - t + \frac{t^{2}}{1 + t}$$ and integrate it to obtain $$\log(1 + x) = \int_{0}^{x}\frac{dt}{1 + x} = x - \frac{x^{2}}{2} + \int_{0}^{x}\frac{t^{2}}{1 + t}\,dt$$ and then analyze the behavior of error term $$\int_{0}^{x}\frac{t^{2}}{1 + t}\,dt$$ for $|x| < 1$ and get the approximation $$\log (1 + x) \approx x - \frac{x^{2}}{2}$$ when $x$ is small enough.
A: First part you can solve it. So we can write according to question $ln(1+x)=x-\frac{x^2}{2}$. Now for the inequality, take $ln$ both the sides and use above formluae. After appying above equation, you will get the inequality like this $$\frac{(n+3)(4n-1)}{4}<(n-1)(2n-1)$$ which will turn into 
$$4n^2-23n+7>0$$ which gives $$n\ge6$$ which is the final answer
Hope this will help !
A: @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.
A: \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 $0$.
