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?

Starting the summations at $k=1$ and using Pochhammer symbols
$$\prod_{n=0}^i(k+n)=k (k+1)_i$$
$$\sum_{k=1}^N\frac 1{k (k+1)_i}=\frac{1}{i \,\Gamma (i+1)}-\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}$$ Now, using Stirling approximation for large values of $N$
$$\log\Bigg[\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\Bigg]=-(i \log (N)+\log (i))-\frac{i (i+1)}{2 N}+O\left(\frac{1}{N^2}\right)$$
$$\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\sim\frac{N^{-i}}{i}\,e^{-\frac{i (i+1)}{2 N}}$$
Edit
Going from $n$ to $2n$ as asked in comments is much more difficult since the result invoke the Gaussian hypergeometric functions.
$$\prod_{n=0}^i(k+2n)=2^i k \left(\frac{k}{2}+1\right)_i$$ and the summation up to $N$ gives
$$S_i=\frac{\sqrt{\pi } 2^{-i-1} \,
   _2F_1\left(\frac{1}{2},1;i+\frac{3}{2};1\right)}{\Gamma
   \left(i+\frac{3}{2}\right)}+\frac{2^{-i-1} \, _2F_1(1,1;i+2;1)}{\Gamma (i+2)}-$$
$$\frac{2^{-i-1} \Gamma \left(\frac{N}{2}+\frac{1}{2}\right) \,
   _2F_1\left(1,\frac{N+1}{2};\frac{1}{2} (2 i+N+3);1\right)}{\Gamma
   \left(i+\frac{N}{2}+\frac{3}{2}\right)}-$$ $$\frac{2^{-i-1} \Gamma
   \left(\frac{N}{2}+1\right) \,
   _2F_1\left(1,\frac{N+2}{2};i+\frac{N}{2}+2;1\right)}{\Gamma
   \left(i+\frac{N}{2}+2\right)}$$ Once expanded, they write in a simple manner in terms of the gamma function.
$$S_1=\frac{1}{4} \left(3-\frac{(N+4) \Gamma \left(\frac{N}{2}+1\right)}{2 \Gamma
   \left(\frac{N}{2}+3\right)}-\frac{(N+3) \Gamma \left(\frac{N+1}{2}\right)}{2
   \Gamma \left(\frac{N}{2}+\frac{5}{2}\right)}\right)$$
$$S_2=\frac{1}{8} \left(\frac{11}{12}-\frac{(N+6) \Gamma \left(\frac{N}{2}+1\right)}{4 \Gamma
   \left(\frac{N}{2}+4\right)}-\frac{(N+5) \Gamma \left(\frac{N+1}{2}\right)}{4
   \Gamma \left(\frac{N}{2}+\frac{7}{2}\right)}\right)$$
$$S_3=\frac{1}{16} \left(\frac{7}{30}-\frac{(N+8) \Gamma \left(\frac{N}{2}+1\right)}{6 \Gamma
   \left(\frac{N}{2}+5\right)}-\frac{(N+7) \Gamma \left(\frac{N+1}{2}\right)}{6
   \Gamma \left(\frac{N}{2}+\frac{9}{2}\right)}\right)$$
$$S_4=\frac{1}{32} \left(\frac{163}{3360}-\frac{(N+10) \Gamma \left(\frac{N}{2}+1\right)}{8 \Gamma
   \left(\frac{N}{2}+6\right)}-\frac{(N+9) \Gamma \left(\frac{N+1}{2}\right)}{8
   \Gamma \left(\frac{N}{2}+\frac{11}{2}\right)}\right)$$ where you can see the simple patterns.
Expanding the gamme functions leads to the expressions you wrote.
When $N\to \infty$, the constant term is
$$T_i=2^{-i-1} \left(\frac{\sqrt{\pi } \,
   _2F_1\left(\frac{1}{2},1;i+\frac{3}{2};1\right)}{\Gamma
   \left(i+\frac{3}{2}\right)}+\frac{\, _2F_1(1,1;i+2;1)}{\Gamma (i+2)}\right)$$ that is to say
$$T_i=\frac 1{2^{i+1}\,i ^2} \left(\frac{i\sqrt{\pi } }{\Gamma \left(i+\frac{1}{2}\right)}+\frac{1}{\Gamma (i)}\right)=\frac 1{2^{i+1}}\frac{1}{i \,\Gamma (i+1)}\left(1+\frac{\sqrt{\pi }\, \Gamma (i+1)}{\Gamma \left(i+\frac{1}{2}\right)}\right)$$ and then
$$S_i\sim T_i-\frac {N^{-i}} i e^{-\frac{i (2 i+1)}{2 N} }$$
