Inequality with logarithmic identity

Question

Prove that $$\frac{1}{3n^2}<\log \left({{n+1} \over {n-1}} \right)^{\frac{n}{2}}-1<\frac{1}{3(n^2-1)}$$ for all $n>2$.

Answer 1

Your equation can be rewritten using $x=1/n$ as $$ \frac{x^2}3<\frac{\ln(1+x)-\ln(1-x)}{2x}-1<\frac{x^2}{3(1-x^2)} $$

Using the logarithmic series for $|x|<1$, i.e., $n>1$, gives \begin{align} \frac{\ln(1+x)-\ln(1-x)}{2x} &=1+\frac{x^2}3+\frac{x^4}5+…+\frac{x^{2k}}{2k+1}+… \\ &=1+\frac{x^2}3\left(1+\frac{3x^2}5+…+\frac{3x^{2k}}{2k+3}+…\right) \end{align}

Using the Bernoulli inequality the coefficients can be bounded by $$(\frac35)^k=\frac1{(1+\frac23)^k}<\frac3{3+2k}<1.$$ This results in the lower bound $$ \frac{x^2}3\left(1+\frac35x^2+\bigl(\frac35\bigr)^2x^4+…\right) =\frac{x^2}{3(1-\frac35x^2)} $$ and the upper bound

$$ \frac{x^2}3(1+x^2+x^4+…) =\frac{x^2}{3(1-x^2)}. $$ Combined they give the stated result $$ \frac{x^2}{3}<\frac{5x^2}{15-3x^2} <\frac{\ln(1+x)-\ln(1-x)}{2x}-1 <\frac{x^2}{3(1-x^2)}. $$

Answer 2

$\ln \left(\left({{n+1} \over {n-1}} \right)^{\frac{n}{2}}\right)=\frac{n}{2}\left(\ln (n+1)-\ln(n-1)\right)$

$\ln (n+1) = \ln n -\sum_\limits{k=1}^{\infty}\frac{(-1)^k}{k(n)^k}\\ \ln (n-1) = \ln n -\sum_\limits{k=1}^{\infty}\frac{(1)^k}{k(n)^k}$

$\frac 2n + \frac 2{3n^3}<\ln (n+1) - \ln(n-1)$

$\frac 1{3n^2}<\ln \left(\left({{n+1} \over {n-1}} \right)^{\frac{n}{2}}\right)-1$