Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$

Question

$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+v)}=\frac{1}{vv!}$$ I am struggling to find a solution for this but no luck yet. How can I analyze it to get to second part?

Answer 1

\begin{eqnarray} \frac{1}{n(n+1)...(n+v)}&=&\frac{1}{v}\frac{(n+v)-n}{n(n+1)...(n+v)}\\ &=&\frac{1}{v}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}] \end{eqnarray} $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)...(n+v)} =\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]$$ $$\frac{1}{v}\sum_{n=1}^{k}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}] =\frac{1}{v}[\frac{1}{v!}-\frac{1}{(k+1)...(k+v)}]$$ Let $k\rightarrow\infty$, then $\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}] =\frac{1}{v}\frac{1}{v!}$

Answer 2

Let $n^{\overline{r}}=\underbrace{n(n+1)(n+2)\cdots (n+r-1)}_{r\text{ terms}}$.

Hence $$\begin{align}\require{cancel} &\sum_{n=1}^{\infty}\frac1{n(n+1)(n+2)\cdots (n+v)}\\ &=\sum_{n=1}^{\infty}\frac1{n(n+1)^{\overline{v}}}=\sum_{n=1}^{\infty}\frac1{n^\overline{v}(n+v)}\\ &=\frac1v\sum_{n=1}^{\infty}\frac1{(n)^\overline{v}}-\frac1{(n+1)^{\overline{v}}}\\ &=\frac1v\left[\left(\frac1{1^\overline{v}}-\cancel{\frac1{2^\overline{v}}}\right)+\left(\cancel{\frac1{2^\overline{v}}}-\bcancel{\frac1{3^\overline{v}}}\right)+\left(\bcancel{\frac1{3^\overline{v}}}-\cancel{\frac1{4^\overline{v}}}\right)+\cdots\right]\\ &=\frac1v \left[\frac1{1^\overline{v}}\right]\\ &=\frac1{vv!}\qquad\blacksquare\end{align}$$