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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?

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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: $$\sum_{n=1}^\infty \dfrac{n!}{b\left(b+1\right)\dots\left(b+n-1\right)}=\dfrac{1}{b-2}\sum_{n=1}^\infty \left(a_n-a_{n+1}\right)=\dfrac{a_1}{b-2}=\boxed{\dfrac{1}{b-2}}$$

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

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