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.
 A: OK, I believe there is.  Here is my result:

$$\displaystyle \sum_{i=1}^n \frac1{\displaystyle \prod_{j=0}^k (i+j)} = \frac1{k \cdot k!} \left [ 1 -  \frac1{\displaystyle\binom{n+k}{k}} \right ] = \frac1{k} \left [\frac1{k!} - \frac1{(n+1)(n+2)\cdots(n+k)}\right ]$$

This result checks out for all of the values of $k$ and $n$ I have plugged into Mathematica.  It reduces to a well-known result in the limit as $n \to \infty$.
The RHS is a result of the following expansion, which may be proven using recursive partial fraction decompositions:
$$\frac1{\displaystyle \prod_{j=0}^k (i+j)} = \frac1{k!} \sum_{j=0}^k (-1)^j \binom{k}{j} \frac1{i+j}$$
By rearranging, we get the more convenient form:
$$\frac1{\displaystyle \prod_{j=0}^k (i+j)} = \frac1{k!} \sum_{j=0}^{k-1} (-1)^j \binom{k-1}{j} \left (\frac1{i+j}-\frac1{i+j+1} \right )$$
Now it is easy to sum over $i$; we get, after rearranging again:
$$\sum_{i=1}^n\frac1{\displaystyle \prod_{j=0}^k (i+j)} = \frac{n}{k \cdot k!} \sum_{j=1}^{k-1} \frac{(-1)^{j-1}}{j+n} \binom{k}{j} $$
This sum is evaluated by defining
$$f(x) = \sum_{j=1}^{k} \frac{(-1)^{j-1}}{j+n} \binom{k}{j} x^{j+n} $$
Then we differentiate and invoke the binomial theorem:
$$f'(x) = x^{n-1} \sum_{j=1}^{k} (-1)^{j-1} \binom{k}{j} x^j  = x^{n-1} \left [ 1-(1-x)^k\right ]$$
We may then conclude that
$$\sum_{j=1}^{k} \frac{(-1)^{j-1}}{j+n} \binom{k}{j} = \int_0^1 dx \, x^{n-1} \left [ 1-(1-x)^k\right ] = \frac1{n} - \frac{(n-1)! k!}{(n+k)!} $$
The stated result follows.
A: 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
$$\sum_{i=1}^n\frac{1}{\prod\limits_{j=0}^{k}(i+j)}
= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(1+j)}-\frac{1}{\prod\limits_{j=0}^{k-1}((n+1)+j)}\right] = \frac{1}{k}\left[\frac{1}{k!} - \frac{n!}{(n+k)!}\right]$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\on{f}\pars{n, k} & \equiv
\bbox[5px,#ffd]{\sum_{i = 1}^{n}
{1 \over \prod_{j = 0}^{k}\,\pars{i + j}}} =
\sum_{i = 1}^{n}{1 \over i^{\overline{k + 1}}} =
\sum_{i = 1}^{n}{1 \over \Gamma\pars{i + k + 1}/\Gamma\pars{i}}
\\[5mm] & =
{1 \over k!}\sum_{i = 1}^{n}{\Gamma\pars{i}\Gamma\pars{k + 1} \over \Gamma\pars{i + k + 1}} =
{1 \over k!}\sum_{i = 1}^{n}
\int_{0}^{1}t^{i - 1}\,\pars{1 - t}^{k}\,\dd t
\\[5mm] & =
{1 \over k!}\int_{0}^{1}\pars{1 - t}^{k}
\sum_{i = 1}^{n}t^{i - 1}\,\,\dd t =
{1 \over k!}\int_{0}^{1}\pars{1 - t}^{k}\,\,
{t^{n} - 1 \over t - 1}\,\,\dd t
\\[5mm] = &\
{1 \over k!}\bracks{%
\int_{0}^{1}\pars{1 - t}^{k - 1}\,\,\dd t -
\int_{0}^{1}t^{n}\pars{1 - t}^{k - 1}\,\,\dd t}
\\[5mm] = &\
{1 \over k!}\bracks{%
{1 \over k} - {\Gamma\pars{n + 1}\Gamma\pars{k} \over \Gamma\pars{n + 1 + k}}} =
{1 \over k!}\bracks{%
{1 \over k} - {n!\pars{k - 1}! \over \pars{n + k}!}}
\\[5mm] = &\
\bbx{{1 \over k}\bracks{{1 \over k!} - {n! \over \pars{n + k}!}}} \\ &
\end{align}
