In this biology textbok I found the following approximation:
$$\prod_{i=1}^{k-1}\left(1-\frac{i}{2N}\right) ≈ 1-\frac{{k \choose 2}}{2N} $$
Can you help me to understand this approximation and help me to understand what assumption are needed for this approximation to be useful.
On
They seemingly use the fact that $$1-\frac{{k \choose 2}}{2N} = 1-\sum_{i=1}^{k-1} \frac i{2N}.$$ If you develop the product you find that expression plus other terms involving products of several factors $i/(2N)$. So they only keep the linear terms. The approximation only holds if you can show that the remaining terms are negligible compared to $1-\frac1{2N}{k \choose 2}$.
You actually get ${{k-1} \choose j} \leq k^j$ terms containing $j$ factors of the form $i/(2N)$ (for $j=2,\ldots,k-1$), each of which is smaller than $k/(2N)$. So the sum of all such terms in bounded above by $\left(\frac{k^2}{2N}\right)^j$. Now if $k^2$ is much smaller than $2N$, then all those terms become negligible compared to the approximation RHS.
Still note that the bounds above are quite rough, but this is more to give an idea of a possible solution.
Let $P$ be the product; then
$$\log{P} = \sum_{i=1}^{k-1} \log{\left (1-\frac{i}{2 N} \right )} $$
When $N$ is sufficiently large, then we may use the approximation $\log{(1-y)} \approx -y$ and get
$$\log{P} \approx -\frac1{2 N} \sum_{i=1}^{k-1} i = -\frac1{2 N} \frac{k (k-1)}{2} = -\frac1{2 N} \binom{k}{2}$$
Thus,
$$P \approx e^{-\frac1{2 N} \binom{k}{2}} \approx 1-\frac1{2 N} \binom{k}{2}$$
The above holds when
$$\frac12 \frac1{4 N^2} \binom{k}{2}^2 $$
is sufficiently small.
For the 1st "sufficient", we need
$$\frac12 \frac1{4 N^2} \sum_{i=1}^{k-1} i^2 = \frac1{8 N^2} \frac{(k-1) k (2 k-1)}{6} \approx \frac{k^3}{24 N^2}$$
or $k \lt C_1 N^{2/3}$, where $C_1$ is a small number. For the second sufficient, we need $k^4/(32 N^2)$ to be small, or $k \lt C_2 \sqrt{N}$. For the estimate to hold overall, we need the smaller of the two estimates, or that $k \lt C_2 \sqrt{N}$.