How to show that this difference of products is $O \left( \frac{1}{n^2} \right) $ Let $k \leq n$. Consider the following difference of products:
$$ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right)$$
For $n=1,2,3$, this is clearly $O \left( \frac{1}{n^2} \right) $ for $k=1,2,3$ respectively. How to show it for arbitrary $n$? And is it possible to give an estimate $C$ such that the difference is less than $\frac{C}{n^2}$?
 A: We have
$$\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)=\left(\prod_{i=1}^{k-2}\,\left(1-\frac{i}{n}\right)\right)\left(\left(1+\frac{1}{n}\right)^{-(k-1)}-\left(1-\frac{k-1}{n}\right)\right)\,.$$
For large $n\in\mathbb{N}$, $\prod_{i=1}^{k-2}\,\left(1-\frac{i}{n}\right)=1+O\left(\frac{1}{n}\right)$ and $$\left(1+\frac{1}{n}\right)^{-(k-1)}=1-\frac{k-1}{n}+\frac{k(k-1)}{2n^2}+O\left(\frac{1}{n^3}\right)\,.$$
Consequently, 
$$\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)=\frac{k(k-1)}{2n^2}+O\left(\frac{1}{n^3}\right)=O\left(\frac{1}{n^2}\right)\,.$$
Indeed, you can easily show that, for $k>1$,
$$\frac{k(k-1)}{2n^2}\left(1-\frac{(k-2)(3k-1)}{6n}\right)\leq \prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)\leq \frac{k(k-1)}{2n^2}\,.$$
A: $$\begin{align}\prod_{i=1}^{k-1}{\left (1-\frac{i}{n+1} \right )} &= \prod_{i=1}^{k-1}{\left (1-\frac{i}{n} + \frac{i}{n^2} \right )} + O \left (\frac1{n^3} \right )\\ &= \prod_{i=1}^{k-1}{\left (1-\frac{i}{n}  \right )} + \sum_{j=1}^{k-1}\frac{j}{n^2} \prod_{i=1,i \ne j}^{k-1}{\left (1-\frac{i}{n}  \right )}+ O \left (\frac1{n^3} \right )\\ &=\prod_{i=1}^{k-1}{\left (1-\frac{i}{n}  \right )} + \sum_{j=1}^{k-1}\frac{j}{n^2}+O \left (\frac1{n^3} \right )\\ &=\prod_{i=1}^{k-1}{\left (1-\frac{i}{n}  \right )} + \frac{k(k-1)}{2 n^2}+O \left (\frac1{n^3} \right )\end{align}$$
Thus the difference is, when $k \lt n$,
$$\frac{k(k-1)}{2 n^2}+O \left (\frac1{n^3} \right )$$
