Summation of the reciprocals of the product of consecutive integers It is well known that there is a closed formula for:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(n)(n + 1)}$$
And likewise for:
$$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots + \frac{1}{(n)(n + 1)(n+2)}$$
I am wondering if there is a closed formula for:
$$f(n, k) = \sum_{i=1}^n \frac{1}{\prod_{j=0}^k (i + j)}$$
Note that putting $k = 1$ and $k = 2$ in the above function yields the above two series.

Notice
$$\begin{align}
\frac{1}{\prod\limits_{j=0}^{k}(i+j)} = \frac{1}{k}\left(\frac{(i+k)-i}{\prod\limits_{j=0}^{k}(i+j)}\right)
&= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_{j=1}^{k}(i+j)}\right]\\
&= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_{j=0}^{k-1}((i+1)+j)}\right]\end{align}$$
The sum can be recasted as a telesoping one. This leads to
$$\sum_{i=1}^n\frac{1}{\prod\limits_{j=0}^{k}(i+j)}
= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(1+j)}-\frac{1}{\prod\limits_{j=0}^{k-1}((n+1)+j)}\right] = \frac{1}{k}\left[\frac{1}{k!} - \frac{n!}{(n+k)!}\right]$$
