Find $\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$

Question

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

Answer 1

\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.

Answer 2

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.

Answer 3

\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 $n = 1$:

$$\frac{1}{k} \frac{1}{1(1+1)...(1+k-1)} = \frac{1}{k * k!}$$