Why is $\left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=1+\frac{x(x-1)}{2j^2}+O(j^{-3})$? On this page, it says
$$ \left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=1+\frac{x(x-1)}{2j^2}+O(j^{-3}). $$
Could anyone please enlighten me how it is deduced? I tried to find an approximation of $j\mapsto \left ( \frac{1}{j}+1 \right )^{x}$, and then multiplying it by $\left ( \frac{x}{j}+1 \right )^{-1}$. But there is no way I could find a such approximation.

As already said, this is the binomial theorem.
You can also do it with Taylor series. Consider first $$A=\left ( \frac{1}{j}+1 \right )^{x}$$ $$\log(A)=x\, \log\left( \frac{1}{j}+1 \right)=x\left(\frac{1}{j}-\frac{1}{2 j^2}+\frac{1}{3 j^3}+O\left(\frac{1}{j^4}\right)\right)$$ $$A=e^{\log(A)}=1+\frac{x}{j}+\frac{x(x-1)}{2 j^2}+\frac{x(x-1)(x-2)}{6
   j^3}+O\left(\frac{1}{j^4}\right)$$ $$B=\left ( \frac{x}{j}+1 \right )^{-1}=\frac{1}{\frac{x}{j}+1}=1-\frac{x}{j}+\frac{x^2}{j^2}-\frac{x^3}{j^3}+O\left(\frac{1}{j^4}\right)$$ All of the above makes $$\left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=A \times B=1+\frac{x(x-1) }{2 j^2}-\frac{(x+1)x(x-1)}{3 j^3}+O\left(\frac{1}{j^4}\right)$$
