# Verifying $\sum\limits_{n=1}^{\infty}\frac{n!}{b(b+1)\cdots(b+n-1)}=\frac{1}{b-2}$

For $$b>2$$, verify that $$\sum\limits_{n=1}^{\infty}\frac{n!}{b(b+1)\cdots(b+n-1)}=\frac{1}{b-2}$$ I am trying to factorize the sum as $$\frac{n!}{b(b+1)\cdots(b+n-2)}-\frac{n!}{b(b+1)\cdots(b+n-1)}$$ From this, how should I proceed further?

Tips:

$$1$$. Find the value of $$\dfrac{1}{b-2}-\dfrac{1}{b}$$

$$2$$. Find the value of $$\dfrac{2}{b\left(b-2\right)}-\dfrac{2}{b\left(b+1\right)}$$

$$3$$. Find the value of $$\dfrac{6}{b\left(b+1\right)\left(b-2\right)}-\dfrac{6}{b\left(b+1\right)\left(b+2\right)}$$

$$4$$. Hence, find the value of $$\dfrac{n!}{b\left(b+1\right)\dots\left(b+n-2\right)\left(b-2\right)}-\dfrac{n!}{b\left(b+1\right)\dots\left(b+n-2\right)\left(b+n-1\right)}$$

$$5$$. By the above questions, find the answer.

Spoiler:

$$1$$.

$$\dfrac{2}{b\left(b-2\right)}$$

$$2$$.

$$\dfrac{6}{b\left(b+1\right)\left(b-2\right)}$$

$$3$$.

$$\dfrac{24}{b\left(b+1\right)\left(b+2\right)\left(b-2\right)}$$

$$4$$.

$$\dfrac{\left(n+1\right)!}{b\left(b+1\right)\dots\left(b+n-1\right)\left(b-2\right)}$$

$$5$$.

Let $$a_1=1$$ and $$a_n=\dfrac{n!}{b\left(b+1\right)\dots\left(b+n-2\right)}$$ for every positive integer $$n\ge 2$$, because of $$\dfrac{n!}{b\left(b+1\right)\dots\left(b+n-1\right)}= \dfrac{n!}{b\left(b+1\right)\dots\left(b+n-2\right)\left(b-2\right)}-\dfrac{\left(n+1\right)!}{b\left(b+1\right)\dots\left(b+n-1\right)\left(b-2\right)}=\dfrac{1}{b-2}\left(a_n-a_{n+1}\right)$$ Therefore, the answer is: $$\sum_{n=1}^\infty \dfrac{n!}{b\left(b+1\right)\dots\left(b+n-1\right)}=\dfrac{1}{b-2}\sum_{n=1}^\infty \left(a_n-a_{n+1}\right)=\dfrac{a_1}{b-2}=\boxed{\dfrac{1}{b-2}}$$

Let $$f(b)=\sum\limits_{n=1}^{\infty}\frac{n!}{(b-1)b(b+1)\ldots(b+n-1)}$$ $$f(b)=\sum\limits_{n=1}^{\infty}\frac{n!}{(b-1)b(b+1)\ldots(b+n-1)}=\sum\limits_{n=1}^{\infty}\frac{n!}{n}(\frac1{(b-1)b(b+1)\ldots(b+n-2)}-\frac1{b(b+1)\ldots(b+n-1)})=\sum\limits_{n=1}^{\infty}(\frac{(n-1)!}{(b-1)b(b+1)\ldots(b+n-2)}-\frac{(n-1)!}{b(b+1)\ldots(b+n-1)})=\frac1{b-1}-\frac1b+f(b)-f(b+1)$$

so we have $$f(b+1)=\frac1{b-1}-\frac1b=\frac1{(b-1)b}$$ or $$f(b)=\frac1{(b-2)(b-1)}$$

So $$\sum\limits_{n=1}^{\infty}\frac{n!}{b(b+1)\ldots(b+n-1)}=(b-1)f(b)=\frac1{b-2}$$