Find the asymptote and the upper bound for $1\big/ \ln\left(\frac{x+1}{x-1}\right)$ as $x\to\infty$.

Question

I need to show that $\left[\ln\left(\frac{x+1}{x-1}\right)\right]^{-1}$ asymptotically approaches to a line from below that I should determine, and hence, show that $\frac{x}{2} > \left[\ln\left(\frac{x+1}{x-1}\right)\right]^{-1}$ for $x>1$. I would be very happy if someone helps. Thank you.

Answer 1

Since $\ln (1+t)<t~\forall~t>0$, \begin{align*} \ln\left(1+\frac2{x-1}\right)&<\frac2{x-1}~\forall~ x>1\\ \frac1{\ln\left(1+\frac2{x-1}\right)}&>\frac{x-1}2~\forall~ x>1\tag{1}\\ \end{align*} Consider $f(x)=\frac x2-\frac1{\ln\left(1+\frac2{x-1}\right)}~\forall~ x>1$. \begin{align*} \lim_{x\to1}f(x)&=\frac12\\ \lim_{x\to\infty}f(x)&=\lim_{x\to\infty}\frac{x\ln\left(1+\frac2{x-1}\right)-2}{2\ln\left(1+\frac2{x-1}\right)}=\lim_{t\to\infty}\frac{(t+1)\ln\left(1+\frac2t\right)-2}{2\ln\left(1+\frac2t\right)}\\&=\lim_{t\to\infty}\frac{(t+1)\left(\frac2t-\frac2{t^2}+O(t^2)\right)-2}{\frac4t}=\lim_{t\to\infty}\frac{(t+1)\left(1-\frac1t\right)-t}{2}=0\\ &\Rightarrow \boxed{\frac1{\ln\left(1+\frac2{x-1}\right)}\to\frac x2\text{ as }x\to\infty,~\forall~ x>1}\\ f^\prime(x)&=\frac12-\frac2{(x-1)^2}\cdot\frac1{\ln^2\left(1+\frac2{x-1}\right)}<\frac12-\frac{2}{(x-1)^2}\cdot \frac{(x-1)^2}{4}=0&(\because (1))\\ &\Rightarrow\boxed{\frac1{\ln\left(1+\frac2{x-1}\right)}<\frac x2~\forall~ x>1}\\ \lim_{x\to\infty}f^\prime(x)&=\frac12-\lim_{t\to\infty}\frac2{t^2\ln^2\left(1+\frac2{t}\right)}=\frac12-\lim_{t\to\infty}\frac2{t^2\cdot\frac4{t^2}}=0\\ \end{align*}

Answer 2

Fix $x > 1$ and write $\varphi(t) = \frac{1}{x+t}$. Since $\varphi_x$ is strictly convex, it follows from the Jensen's inequality that

\begin{align*} \frac{1}{2} \log\left(\frac{x+1}{x-1}\right) = \frac{1}{2} \int_{-1}^{1} \varphi_x(t) \, \mathrm{d}t > \varphi_x \left( \frac{1}{2} \int_{-1}^{1} t \, \mathrm{d}t \right) = \frac{1}{x} \end{align*}

On the other hand, again by the strict convexity of $\varphi_x$,

\begin{align*} \frac{1}{2} \log\left(\frac{x+1}{x-1}\right) = \frac{1}{2} \int_{-1}^{1} \varphi_x(t) \, \mathrm{d}t < \frac{\varphi_x(-1) + \varphi_x(1)}{2} = \frac{x}{x^2-1}. \end{align*}

Altogether, it follows that

$$ \frac{x}{2} - \frac{1}{2x} < \left[\log\left(\frac{x+1}{x-1}\right)\right]^{-1} < \frac{x}{2} $$

for all $x > 1$.

---

Addendum. It is easy to check that the above function admits the Laurent expansion of the form

$$ \left[\log\left(\frac{x+1}{x-1}\right)\right]^{-1} = \frac{x}{2} - \sum_{n=0}^{\infty} \frac{a_{2n+1}}{x^{2n+1}} $$

as $x\to\infty$. Now here comes a hard part: $\texttt{Mathematica}$ seems to suggests $a_{2n+1} > 0$ for all $n \geq 0$. For instance,

$$ a_1 = \frac{1}{6}, \quad a_3 = \frac{2}{45}, \quad a_5 = \frac{22}{945}, \quad a_7 = \frac{214}{14175}, \quad a_9 = \frac{5098}{467775}, \quad \cdots. $$

It will be interesting to be able to prove it, although I have no good idea to begin with.