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For $b \gt 2$ , verify that $$\sum_{n=1}^{\infty}\frac{n!}{b(b+1)...(b+n-1)}=\frac{1}{b-2}$$

This is how I tried.. $$\sum_{n=1}^{\infty}\frac{n!(b-2)}{(b-2)b(b+1)...(b+n-1)}=\sum_{n=1}^{\infty}\frac{n!(b+n-1-2+1-n)}{(b-2)b(b+1)...(b+n-1)}=\sum_{n=1}^{\infty}\frac{n!(b+n-1)-(1+n)}{(b-2)b(b+1)...(b+n-1)}=\sum_{n=1}^{\infty}\frac{n!(b+n-1)}{(b-2)b(b+1)...(b+n-1)}-\frac{n!(1+n)}{(b-2)b(b+1)...(b+n-1)}=\frac{1}{b-2}\left(\frac{b-2}{b}+\sum_{n=2}^{\infty}\frac{n!}{b(b+1)...(b+n-2)}-\frac{(n+1)!}{b(b+1)....(b+n-1)}\right)$$

Now $$S_n=\frac{1}{b-2}\left[\frac{b-2}{b}+\left(\frac{2}{b}-\frac{3!}{b(b+1)}\right)+\left(\frac{3!}{b(b+1)}-\frac{4!}{b(b+1)(b+2)}\right)+...+\left(\frac{n!}{b(b+1)..(b+n-2)}-\frac{(n+1)!}{b(b+1)...(b+n-1)}\right)\right]=\frac{1}{b-2}\left[1-\frac{(n+1)!}{b(b+1)..(b+n-1)}\right]$$

All I need to show is that $$\lim\limits_{n\to \infty}\frac{(n+1)!}{b(b+1)..(b+n-1)}=0$$ which I am unable to show.

Thanks for the help!!

Hakim
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tattwamasi amrutam
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3 Answers3

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{n! \over b\pars{b + 1}\ldots\pars{b + n - 1}} ={1 \over b-2}}$

\begin{align} \mbox{Note that}\quad \Gamma\pars{b}&={\Gamma\pars{b + 1} \over b} ={\Gamma\pars{b + 2} \over b\pars{b + 1}} ={\Gamma\pars{b + 3} \over b\pars{b + 1}\pars{b + 2}}=\cdots \\[3mm]&={\Gamma\pars{b + n} \over b\pars{b + 1}\pars{b + 2}\ldots\pars{b + n - 1}} \end{align} such that

\begin{align}&\color{#44f}{\large% \sum_{n = 1}^{\infty}{n! \over b\pars{b + 1}\ldots\pars{b + n - 1}}} =\sum_{n = 1}^{\infty}{\Gamma\pars{n + 1}\Gamma\pars{b} \over \Gamma\pars{b + n}} =\sum_{n = 1}^{\infty}n\,{\Gamma\pars{n}\Gamma\pars{b} \over \Gamma\pars{b + n}} \\[3mm]&=\sum_{n = 1}^{\infty}n\,\int_{0}^{1}t^{n - 1}\pars{1 - t}^{b - 1}\,\dd t =\int_{0}^{1}\ \overbrace{\pars{\sum_{n = 1}^{\infty}nt^{n - 1}}}^{\ds{=\ \pars{1 - t}^{-2}}} \ \pars{1 - t}^{b - 1}\,\dd t =\int_{0}^{1}\pars{1 - t}^{b - 3}\,\dd t \\[3mm]&=\left.-\,{\pars{1 - t}^{b - 2} \over b - 2}\right\vert_{0}^{1} =\color{#44f}{\large{1 \over b - 2}}\,,\qquad\qquad b > 2 \end{align}

$\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$ and we used well known properties of it.

Felix Marin
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$$\lim_{n\to\infty}\frac{(n+1)!}{b(b+1)\cdot...\cdot(b+n-1)}=\lim_{n\to\infty}\frac{n!n^{b}}{b(b+1)\cdot...\cdot(b+n-1)(b+n)}\cdot\frac{(n+1)(b+n)}{n^{b}}$$

$$=\lim_{n\to\infty}\frac{n!n^{b}}{b(b+1)\cdot...\cdot(b+n)}\cdot\frac{(1+\frac{1}{n})(1+\frac{b}{n})}{n^{b-2}}=\Gamma(b)\lim_{n\to\infty}\frac{(1+\frac{1}{n})(1+\frac{b}{n})}{n^{b-2}}=0$$

where I have used that $b>2$ to justify the existence of $$\lim_{n\to\infty}\frac{(1+\frac{1}{n})(1+\frac{b}{n})}{n^{b-2}}$$

user71352
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$b = 3 \Rightarrow \sum\limits_{n = 1}^\infty {\frac{{n!}}{{3 \times 4 \times ....\left( {n - 2} \right)}}} = \sum\limits_{n = 1}^\infty {\left( {2n\left( {n - 1} \right)} \right)} = + \infty $

sabaga
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