# Deduce that $e < \left({1+ \frac 1n}\right)^{n+\frac12} < e^{1+\frac {1}{2n (n+1)}}$ from the result obtained.

Prove that $$2x < \log{\frac {1+x}{1-x}} < 2x \left[{\frac {1+x^2}{3 (1-x^2)}}\right]$$ where $$0 < x <1.$$ Hence, deduce that $$e < \left({1+ \frac 1n}\right)^{n+\frac12} < e^{1+\frac {1}{2n (n+1)}}$$

I have solved the first part by taking derivative but I'm not able to deduce the second part from the result obtained.

It is simple. Just substitute $$x$$ with $$\frac{1}{1+2n}$$. Then take exponential on both sides.
To prove RHS, you might need an extra inequality that: $$1< 1+2n$$. Which is true only because $$x < 1$$ and $$x > 0$$.
The second part is equivalent to $$\frac{2}{2n+1}<\ln(1+\tfrac1n)<1+\tfrac{1}{2n(n+1)}$$. The first of these expressions is $$2x$$ with $$x=\tfrac{1}{2n+1}$$, as @Inuyashayagami proposed. With this choice of $$x$$,$$n=\frac{1-x}{2x}\implies\ln\left(1+\tfrac1n\right)=\ln\left(\tfrac{1+x}{1-x}\right),\,1+\tfrac{1}{2n(n+1)}=\tfrac{1+x^2}{1-x^2}.$$Now we just need to check $$\frac{2x(1+x^2)}{3(1-x^2)}\le\tfrac{1+x^2}{1-x^2}$$, i.e. $$\frac{2x}{3}\le1$$, which is trivial.