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

Question

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.

Answer 1

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 $x > 0$.

Answer 2

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.