How and why does $\frac{(k+1)(c+k)}{(a+k)(b+k)}=1+\frac{1+c-a-b}{k}+O\left(\frac{1}{k^2}\right)$ hold for integer $k$ and complex $a,b,c$?

Question

The question is rather simple:

How and why does this identity hold? $$\frac{(k+1)(c+k)}{(a+k)(b+k)}=1+\frac{1+c-a-b}{k}+O\left(\frac{1}{k^2}\right)$$ where $k$ is an integer and $c,a,b$ are complex.

This is in the context of hypergeometric functions, should it matter.

I've tried partial fractions and different equivalent expressions but cannot get to obtain the result. Any tips or pointers would be much appreciated.

Answer 1

Note that we have

$$\begin{align} \frac{(k+1)(k+c)}{(k+a)(k+b)}&=\frac{k^2+(c+1)k+c}{k^2+(a+b)k+ab}\\\\ &=\frac{1+\frac{c+1}{k}+\frac{c}{k^2}}{1+\frac{a+b}{k}+\frac{ab}{k^2}}\\\\ &=\left(1+\frac{c+1}{k}+\frac{c}{k^2}\right)\left(1-\frac{a+b}{k}-\frac{ab}{k^2}+O\left(\frac1{k^2}\right)\right)\\\\ &=1+\frac{c+1-a-b}{k}+O\left(\frac1{k^2}\right) \end{align}$$

as was to be shown!

Answer 2

Step 1, divide each parenthesis by $k$. This yields: $$\frac{(1+1/k)(1+c/k)}{(1+a/k)(1+b/k)}$$ For small $1/k$, we use $$\frac{1}{1+a/k}=1-\frac ak+O\left(\frac 1{k^2}\right)$$ Next, in the product, just keep the terms in $(1/k)^0$ and $(1/k)$: $$\left(1+\frac 1k\right)\left(1+\frac ck\right)\left(1-\frac ak+O\left(\frac 1{k^2}\right)\right)\left(1-\frac bk+O\left(\frac 1{k^2}\right)\right)=1+\frac{1+c-a-b}k+O\left(\frac 1{k^2}\right)$$