Sum of infinite series $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}$ Lets define the following series
$$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$
I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc..
the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and  can be found here that is equal to $n*n!$
Finally I want to prove the general equality:
$\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
 A: Hint: Telescopic sum and
$$\frac{n}{i(i+1)(i+2)\cdots(i+n)}=\frac{1}{i(i+1)\cdots(i+n-1)}-\frac{1}{(i+1)(i+2)\cdots(i+n)}.$$
A: If you failed to see the telescoping nature of the series, here is an alternative approach that makes use of the Beta function.
\begin{align*}
S_n &= \sum_{i = 1}^\infty \frac{1}{i (i + 1) (i + 2) \cdots (i + n)} \quad n \in \mathbb{N}\\
&= \sum_{i = 1}^\infty \frac{1 \cdot 2 \cdots (i - 1)}{1 \cdot 2 \cdots (i - 1)i(i + 1) \cdots (i + n)}\\
&= \sum_{i = 1}^\infty \frac{(i - 1)!}{(i + n)!}\\
&= \sum_{i = 1}^\infty \frac{\Gamma (i)}{\Gamma (i + n + 1)}\\
&= \frac{1}{\Gamma (n + 1)} \sum_{i = 1}^\infty \frac{\Gamma (i) \Gamma (n + 1)}{\Gamma (i + n + 1)}\\
&=\frac{1}{\Gamma (n + 1)} \sum_{i = 1}^\infty \text{B}(i, n + 1)\\
&=\frac{1}{\Gamma (n + 1)} \sum_{i = 1}^\infty \int_0^1 t^{i - 1} ( 1 - t)^n \, dt\\
&=\frac{1}{\Gamma (n + 1)} \int_0^1 (1 - t)^n \left [\sum_{i = 1}^\infty t^{i - 1} \right ] \, dt\\
&= \frac{1}{\Gamma (n + 1)} \int^1_0 (1 - t)^n \cdot \frac{1}{1 - t} \, dt\\
&= \frac{1}{\Gamma (n + 1)} \int^1_0 (1 - t)^{n - 1} \, dt\\
&= \frac{1}{\Gamma (n + 1)} \left [-\frac{1}{n} (1 - t)^n \right ]^1_0\\
&= \frac{1}{\Gamma (n + 1) n}\\
&= \frac{1}{n \cdot n!},
\end{align*}
as expected.
A: $$\frac{1}{i(i+1)(i+2)\cdots(i+n)}$$
$$=\frac{1}{n}\left(\frac{1}{i(i+1)\cdots(i+n-1)}-\frac{1}{(i+1)(i+2)\cdots(i+n)}.\right)$$ 
The first term of this series becomes 
$$=\frac{1}{n}\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right) $$
The second term becomes 
$$=\frac{1}{n}\left(\frac{1}{(n+1)!}-\frac{2}{(n+2)!}\right) $$
The third term becomes 
$$=\frac{1}{n}\left(\frac{2}{(n+2)!}-\frac{6}{(n+3)!}\right)$$
Hence we see that the sum telescopes to $\frac {1}{n. n!}$
A: I will add another answer with a different approach (less elementary but also interesting). From finite calculus theory your series can also be stated as $$\sum_{k=1}^\infty\frac1{k^\overline n}=\frac1{1^\overline n}+\sum_{k=2}^\infty\frac1{k^\overline n}=\frac1{n!}+\sum\nolimits_2^\infty (k-1)^\underline{-n}\delta k$$ where $a^\overline c$ is a rising factorial and $a^\underline c$ is a falling factorial. Hence 
$$\begin{align}\sum_{k=1}^\infty\frac1{k^\overline n}&=\frac1{n!}+\frac{(k-1)^\underline{1-n}}{1-n}\bigg|_{k=2}^{k\to\infty}\\&=\frac1{n!}+0-\frac{1^\underline{1-n}}{1-n}\\&=\frac1{n!}+\frac1{(n-1)2^\overline{n-1}}\\&=\frac1{n!}+\frac1{(n-1)n!}\\
&=\frac1{n!\, n}\end{align}$$
because $\frac{a^\underline m}{m}$ is a primitive (in the finite calculus sense) of $a^\underline {m-1}$, in the domain where both expressions are well-defined.
