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?

I think I've posted this before,
but what the heck.
The telescoping occurs not only
for the product of consecutive integers
but for its reciprocal.
Here's the proofs.
$p_m(x)
=\prod_{k=0}^{m-1} (x+k)
$.
$\begin{array}\\
p_m(x+1)-p_m(x)
&=\prod_{k=0}^{m-1} (x+1+k)-\prod_{k=0}^{m-1} (x+k)\\
&=\prod_{k=1}^{m} (x+k)-\prod_{k=0}^{m-1} (x+k)\\
&=(x+m-x)\prod_{k=1}^{m-1} (x+k)\\
&=m\prod_{k=0}^{m-2} (x+1+k)\\
&=mp_{m-1}(x+1)\\
p_m(x)-p_m(x-1)
&=mp_{m-1}(x)\\
\dfrac1{p_m(x)}-\dfrac1{p_m(x+1)}
&=\dfrac1{\prod_{k=0}^{m-1} (x+k)}-\dfrac1{\prod_{k=0}^{m-1} (x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{m-1} (x+k)}-\dfrac1{\prod_{k=1}^{m} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{m-1} (x+k)}(\dfrac1{x}-\dfrac1{x+m})\\
&=\dfrac1{\prod_{k=1}^{m-1} (x+k)}(\dfrac{x+m-x}{x(x+m)})\\
&=\dfrac{m}{\prod_{k=0}^{m} (x+k)}\\
&=\dfrac{m}{p_{m+1}(x)}\\
\end{array}
$
$\begin{array}\\
q_m(n)
&=\sum_{x=1}^{n} p_{m}(x)\\
&=\dfrac1{m+1}\sum_{x=1}^{n} (p_{m+1}(x)-p_{m+1}(x-1))\\
&=\dfrac{p_{m+1}(n)}{m+1}\\
r_m(n)
&=\sum_{x=n}^{\infty}\dfrac1{p_{m}(x)}\\
&=\dfrac1{m-1}\sum_{x=n}^{\infty}\left(\dfrac1{p_{m-1}(x)}-\dfrac1{p_{m-1}(x+1)}\right)\\
&=\dfrac1{(m-1)p_{m-1}(n)}\\
\end{array}
$
