Compute approximation difference

Question

How can one compute

$$e^{\frac{-j(j+1)}{2n}} -\prod_{i=1}^{j}\left(1-\frac{i}{n} \right)$$

assuming $j,n \geq 2$? I am interested in understanding how quickly this number tends towards zero as $n$ grows.

Answer 1

The first order difference is $\frac{j(j+1)(2j+1)}{6n^2}$

The Taylor expansion of the exponential to order $n^{-2}$ is $1-\frac {j(j+1)}{2n}+\frac {j^2(j+1)^2}{8n^2}$

If we expand the product it is $(1-\frac 1n)(1-\frac 2n)\ldots (1-\frac jn)$ We get the $1$ by taking the first term in each factor. The $\frac 1n$ term comes from taking one of the $\frac kn$ terms and all the rest $1$s in the product. Adding them all up we get $-\frac {j(j+1)}{2n}$. To get a term of order $\frac 1{n^2}$ we choose two terms to take the $\frac kn$ term. Adding up all the pairs is $\frac 1{2n^2} (1+2+3+\ldots +j)^2-\frac 1{n^2}(1^2+2^2+3^2+\ldots +j^2)=\frac {j^2(j+1)^2}{8n^2}-\frac {j(j+1)(2n+1}{6n^2}$ and the last term is the asserted difference.