Series expansion of $\left(\frac{x+1}{x-1}\right)^x$ at $x=\infty$

Question

Find the series expansion of$$\left(\frac{x+1}{x-1}\right)^x$$ at $x=\infty$

I'm not particularly familiar with computation of series at infinity. Moreover the derivatives are hard to find. Any hint is appreciated.

Here is wolframalpha's answer.

Answer 1

You can obtain the expansion by using the Maclaurin series of $\log(1\pm w)$ about $w=0$ with $w=1/x$: \begin{align*} \left( {\frac{{x + 1}}{{x - 1}}} \right)^x & = \exp \left( {x\log \left( {\frac{{1 + \frac{1}{x}}}{{1 - \frac{1}{x}}}} \right)} \right) \\&= \exp \left( {2x\left( {\frac{1}{x} + \frac{1}{{3x^3 }} + \frac{1}{{5x^5 }} + \cdots } \right)} \right) \\ & = e^2 \exp \left( {\frac{2}{{3x^2 }} + \frac{2}{{5x^4 }} + \cdots } \right) \\ & = e^2 + \frac{{2e^2 }}{{3x^2 }} + \frac{{28e^2 }}{{45x^4 }} + \cdots \, . \end{align*} In the last step I composed the series in the argument of the exponential with the Maclaurin series of the exponential function and expanded in inverse powers of $x$.