show $\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^{x}=1+\frac{x(x-1)}{2j^{2}}+O(\frac{1}{j^{3}})$ I should show the following equality provided $x\in\mathbb{C}$ without negative integers:
$$\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^{x}=1+\frac{x(x-1)}{2j^{2}}+O(\frac{1}{j^{3}})$$
I tried using binomial theorem, then I get for $(1+\frac{1}{j})^{x} = 1+ \frac{x}{j} +\frac{x(x-1)}{2j^{2}} +O(\frac{1}{j^{3}}) $. So it seems to go in the right direction but I don't know how multiplying with $\left(1+\frac{x}{j}\right)^{-1}$ brings me to the right result. Any help is greatly appreciated.
Geometrically &amp; binomially expand ...
\begin{eqnarray*}
\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^{x}= \left(1-\frac{x}{j}+\frac{x^2}{j^2}+ O(\frac{1}{j^{3}}) \right)\left(1+\frac{x}{j}+\frac{x(x-1)}{2} \frac{1}{j^2} +O(\frac{1}{j^{3}}) \right)
\end{eqnarray*}
okay so if we have some $x\in\mathbb{C}$ then it is fair to write it as $x=a+bj$
now:
$$\frac1j=\frac1j \frac jj=-j$$ and so:
$$\frac xj=-j(a+bj)=-aj+b$$
see if you can do anything with this

looking at the other part:
$$\left(1+\frac1j\right)^x=\left(1-j\right)^x=\left(\sqrt{2}e^{7\pi j/4}\right)^x$$
Now i suggest multiplying this term by the expansion of $(1-xj)^{-1}$ and see what you get