Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)}=\frac{1}{3\cdot3!}-\frac{1}{3\cdot(N+1)(N+2)(N+3)}$$
The pattern is $$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)(N+2)(N+3)\cdot\cdot\cdot\cdot(N+i)}$$
which is easy to prove by induction.
As an easy consequence it follows that $$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}$$
I did not find this summation of reciprocal products in my mathbooks. Is this a known theorem?
Can anyone point me in the right direction for further study?

This identity is a matter of telescoping. Let $q$ be a positive integer.

We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q)}\left((k+q)-k\right)\tag{1.1}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)}
-\frac{1}{q}\sum_{k=1}^N\frac{1}{(k+1)\cdots(k+q)}\tag{1.2}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)}\\
&\qquad\quad-\frac{1}{q}\sum_{k=2}^{N+1}\frac{1}{k(k+1)\cdots(k+q-1)}\tag{1.3}\\
&=\frac{1}{q}\left(\frac{1}{q!}-\frac{1}{(N+1)(N+2)\cdots(N+q)}\right)\tag{1.4}\\
&\,\,\color{blue}{=\frac{1}{q}\left(\frac{1}{q!}-\frac{N!}{(N+q)!}\right)}\tag{1.5}\\
\end{align*}
and the claim follows.

Comment:

*

*In (1.1) we expand numerator and denominator with $q$.


*In (1.2) we split the sums and cancel terms accordingly.


*In (1.3) we shift the index of the right-hand sum and start with $k=2$.


*In (1.4) we do the cancellation thanks to telescoping.
Note: A theory behind it is the calculus of finite differences. In fact this identity is an application of a discrete analogon of the fundamental theorem of calculus. An application of it is given for instance here.
Calculus of finite differences: Let's do the calculation from above a bit more sophisticated.

*

*In terms of calculus of finite differences we consider the falling factorial operator
\begin{align*}
x^{\underline{q}}=x(x-1)\cdots(x-q+1)
\end{align*}


*which can be expanded to negative integers, written as $-q$ with $q> 0$ which is then defined as
\begin{align*}
\color{blue}{x^{\underline{-q}}:=\frac{1}{(x+1)(x+2)\cdots(x+q)}}\tag{2.1}
\end{align*}


*The forward difference operator $\Delta$ defined as
\begin{align*}
\left(\Delta F\right)(x):=F(x+1)-F(x)
\end{align*}
applied to $x^{\underline{-q}}$ gives
\begin{align*}
\color{blue}{\Delta x^{\underline{-q}}=-qx^{\underline{-q-1}}}\tag{2.2}
\end{align*}


*A discrete analogon to the fundamental theorem of calculus is given as
\begin{align*}
\color{blue}{\sum_{k=0}^{N-1}\left(\Delta F\right)(k)}&\color{blue}{=F(N)-F(0)}\tag{2.3}\\
\int_{0}^xF^{\prime}(t)\,dt&=F(x)-F(0)
\end{align*}

We can now perform the calculation from above as
\begin{align*}
\color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}}
=\sum_{k=1}^{N}(k-1)^{\underline{-\left(q+1\right)}}\tag{$\to\ $2.1}\\
&=\sum_{k=0}^{N-1}k^{\underline{-\left(q+1\right)}}\tag{shift $k$}\\
&=-\frac{1}{q}\sum_{k=0}^{N-1}\left(\Delta k^{\underline{-q}}\right)\tag{$\to\ $2.2}\\
&=-\frac{1}{q}\left(N^{\underline{-q}}-0^{\underline{-q}}\right)\tag{$\to\ $2.3}\\
&=-\frac{1}{q}\left((N+1)(N+2)\cdots(N+q)-1\cdot 2\cdots q\right)\tag{$\to\ $2.1}\\
&\,\,\color{blue}{=\frac{1}{q}\left(\frac{1}{q!}-\frac{N!}{(N+q)!}\right)}
\end{align*}
and the claim follows again in accordance with the result (1.5).

Note: A generalisation of this identity even more compactly derived is given as formula (1.16) in Differenzen und Summen in Konkrete Analysis by J. Cigler.
