I came across this approximation in the book Principles of Population Genetics by Hartl and Clark (page 130).
$$\prod_{i=1}^{k-1}\left(1-\frac{i}{2N}\right) \approx 1 - \frac{{k \choose 2}}{2N}$$
This approximation is supposed to hold true for large values of $N$. $N$ and $k$ are both positive integers. While the book does not explicitly say that $k<<N$, this assumption would also be reasonable.
Can you help me to prove this approximation right?
Let's assume that $k \ll N$. By using the Taylor series expansion $\ln(1-x) = -x+O(x^2)$, we have:
$$\displaystyle\sum_{i = 1}^{k-1}\ln\left(1-\dfrac{i}{2N}\right) \approx \displaystyle\sum_{i = 1}^{k-1}-\dfrac{i}{2N} = -\dfrac{1}{2N}\cdot\dfrac{(k-1)k}{2} = -\dfrac{\binom{k}{2}}{2N}.$$
Then, by exponentiating both sides, and using the Taylor series $e^{-x} = 1-x + O(x^2)$, we have: $$\prod_{i = 1}^{k-1}\left(1-\dfrac{i}{2N}\right) \approx \exp\left[-\dfrac{\binom{k}{2}}{2N}\right] \approx 1-\dfrac{\binom{k}{2}}{2N}.$$