Find $\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$ Original question is $$\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$$
I got it down to $$\sum_{n=1}^{\infty} \frac{(n-1)!}{(k+n)!}$$
Here I am confused.  Possible fraction decomposition but its ugly!  Maybe this approach is not good?  Ideas?
Answer is $$\frac{1}{k \cdot k!}$$
I want to know how to proceed with my work though
 A: \begin{align*}\sum_{n=1}^{\infty}\frac{(n-1)!}{(k+n)!}
&= \frac{1}{k!}\sum_{n=1}^{\infty}\frac{k!(n-1)!}{(k+n)!}\\
&=\frac{1}{k!}\sum_{n=1}^{\infty} \beta(k+1,n)\\
&=\frac{1}{k!}\sum_{n=1}^{\infty} \int_0^1 t^k(1-t)^{n-1}dt\\
&=\frac{1}{k!}\int_0^1 t^k\bigg(\sum_{n=1}^{\infty}(1-t)^{n-1}\bigg)dt\\
&=\frac{1}{k!}\int_0^1 \frac{t^k}{t}dt\\
&=\frac{1}{k \cdot k!}
\end{align*}
Here, $\beta(\cdot,\cdot)$ is beta function.
A: For $k\ge 0$, and $n\ge 1$, let
$$
A_k(n)=\frac{1}{\prod\limits_{i=0}^{k}(n+i)}\ .
$$
Then
$$
A_k(n+1)-A_k(n)=\frac{1}{\prod\limits_{i=0}^{k}(n+1+i)}-
\frac{1}{\prod\limits_{i=0}^{k}(n+i)}
$$
$$
=\frac{1}{\prod\limits_{i=1}^{k+1}(n+i)}-\frac{1}{\prod\limits_{i=0}^{k}(n+i)}
$$
by shifting the index in the first product. Then by factoring out the common factors
$$
A_k(n+1)-A_k(n)=\frac{1}{\prod\limits_{i=1}^{k}(n+i)}
\times\left[\frac{1}{n+k+1}-\frac{1}{n}\right]
$$
$$
=\frac{1}{\prod\limits_{i=1}^{k}(n+i)}
\times\left[\frac{-(k+1)}{n(n+k+1)}\right]\ .
$$
So
$$
A_k(n+1)-A_k(n)=-(k+1)A_{k+1}(n)\ .
$$
Now the wanted series can be computed by telescoping, for $k\ge 1$,
$$
\sum\limits_{n=1}^{\infty}A_k(n)=\frac{1}{k}\sum\limits_{n=1}^{\infty}\left[
A_{k-1}(n)-A_{k-1}(n+1)
\right]=\frac{A_{k-1}(1)}{k}=\frac{1}{k\times k!}\ .
$$
Remark:
The key identity $A_{k-1}(n+1)-A_{k-1}(n)=-k A_{k+1}(n)$ is the discrete analogue of $\frac{d}{dx}x^{-k}=-k x^{-k-1}$. The same kind of argument also works for the products in the numerators. This actually gives a way of computing sums of powers $\sum_{n=1}^{N}n^k$, by changing the linear basis to that of rising powers. This involves the Stirling numbers.
A: \begin{align*}
\sum_{n=1}^\infty \frac{1}{n(n+1)...(n+k)} &= \frac{1}{k} \sum_{n=1}^\infty \frac{k}{n(n+1)...(n+k)} \\
 &= \frac{1}{k} \sum_{n=1}^\infty \left[ \frac{1}{n(n+1)...(n+k-1)} - \frac{1}{(n+1)...(n+k)} \right], \\
\end{align*}
and this series telescopes so that every minus cancels with a plus, and we are left with only the first plus term, when $n = 1$:
$$\frac{1}{k} \frac{1}{1(1+1)...(1+k-1)} = \frac{1}{k * k!}$$
