real analysis - Proof that $prod_{n=1}^N(1 - K/n)$ is $O(N^{-K})$

I am trying to prove Raabe's test and one of the steps is to show that $$\prod_{n=1}^N(1 - K/n)$$ is order $O(N^{-K})$. The book I'm using doesn't give me tools to do this so I'm wondering how it's done.

There's another proof I've found online that involves telescoping sums. However, I am interested in seeing how this proof is done.

Thanks.

Answer

If $0 < \frac{K}{n} < 1$ then we have $1 - \frac{K}{n} = e^{\log\left(1 -\frac{K}{n}\right)} = e^{-\frac{K}{n} - \frac{K^2}{2n^2} - \mathcal{O}\left(\frac{K}{n}\right)^3}$ where we used $\log(1-x) = -x - \frac{x^2}{2} + \mathcal{O}(x^3)$. Fix a $M > \lceil K \rceil$ then if $N > M$ we have

$$\prod_{n=1}^N\left(1 - \frac{K}{n}\right) = \prod_{n=1}^{M-1}\left(1 - \frac{K}{n}\right)\times \prod_{n=M}^N\left(1 - \frac{K}{n}\right) \\= \prod_{n=1}^{M-1}\left(1 - \frac{K}{n}\right)\times \prod_{n=M}^{N}e^{-\frac{K}{n} - \frac{K^2}{2n^2} - \mathcal{O}\left(\frac{K}{n}\right)^3} = \prod_{n=1}^{M-1}\left(1 - \frac{K}{n}\right) \times e^{-K\sum_{n=M}^N \frac{1}{n} + \mathcal{O}(1)}$$

The first product is independent of $N$ so it's just a constant $C$ and using the asymptotics for the harmonic number $\sum_{n=1}^N \frac{1}{n} = \log(n) + \gamma + \mathcal{O}(N^{-1})$ we have $-K\sum_{n=M}^N \frac{1}{n} = -K\log(N) + \mathcal{O}(1)$ and it follows that

$$\prod_{n=1}^N\left(1 - \frac{K}{n}\right) = C e^{-K\log(N) + \mathcal{O}(1)} = \mathcal{O}(N^{-K})~~~\text{as}~~N\to\infty$$