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Compute the sum $$ \sum_{n=1}^\infty \frac{(n-1)!}{(n+k)!} $$ where $k = 1,2,\ldots$ is fixed.

I was able to compute the sum of a simpler infinite series: $$ \sum_{n=1}^\infty \frac 1{n(n+1)(n+2)} $$ In that case, my given hint was to notice that $$ \sum_{n=1}^\infty \frac 2{n(n+1)(n+2)} = \sum_{n=1}^\infty \left(\frac 1{n(n+1)}-\frac 1{(n+1)(n+2)} \right) $$ is a telescoping series.

How can I find a telescoping series for the generalized sum above?

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  • $\begingroup$ Hint: look at what terms "vanish" from the denominator when you go from the original series to the telescoping one. If you are still stuck, try doing it for $k=3$. $\endgroup$ – theindigamer Oct 7 '16 at 2:48
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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}\quad}$ where $\ds{\quad k = 1,2,\ldots}$ is fixed.

Lets $\ds{\,\mrm{f}\pars{x} \equiv \sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}\,x^{n}\quad}$ where $\ds{\quad x \in \pars{0,1}\quad}$ and $\ds{\color{#f00}{\quad\sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}} = \mrm{f}\pars{1^{-}}}$.


\begin{align} x\,\mrm{f}'\pars{x} &\equiv \sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}\,n\,x^{n} = \sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k - 1}!}\,\,x^{n} - k\sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}\,\,x^{n} \\[5mm] & = \sum_{n = 0}^{\infty}{n! \over \pars{n + k}!}\,\,x^{n + 1} - k\,\mrm{f}\pars{x} = {x \over k!} + x^{2}\sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}\,n\,x^{n - 1} -k\,\mrm{f}\pars{x} \\[5mm] & = {x \over k!} + x^{2}\,\mrm{f}'\pars{x} -k\,\mrm{f}\pars{x} \implies \pars{x^{2} - x}\mrm{f}'\pars{x} - k\,\mrm{f}\pars{x} = -\,{x \over k!} \\[5mm] \implies & \bbox[8px,border:0.1em groove navy]{ \mrm{f}'\pars{x} - {k \over x^{2} - x}\,\mrm{f}\pars{x} = {1 \over k!}\,{1 \over 1 - x}\,,\quad\mrm{f}\pars{0^{+}} = 0} \end{align}
The above differential equation can be rewritten as: \begin{align} &\totald{\bracks{x^{k}\pars{1 - x}^{-k}\,\mrm{f}\pars{x}}}{x} = {1 \over k!}\,\pars{1 - x}^{-k - 1}x^{k} \\[5mm] &\ \implies\bbox[8px,border:0.1em groove navy]{ \mrm{f}\pars{x} = {1 \over k!}\,x^{-k}\pars{1 - x}^{k} \int_{0}^{x}\pars{1 - t}^{-k - 1}t^{k}\,\dd t\,,\qquad x \in \pars{0,1}} \end{align}
\begin{align} \color{#f00}{\quad\sum_{n = 1}^{\infty}{\pars{n - 1}! \over \pars{n + k}!}} & = \lim_{x \to 1^{-}}\bracks{% {1 \over k!}\,x^{-k}\pars{1 - x}^{k}\int_{0}^{x}\pars{1 - t}^{-k - 1}t^{k} \,\dd t} = {1 \over k!}\,\lim_{x \to 1^{-}} \bracks{{\pars{1 - x}^{-k - 1}x^{k} \over k\pars{1 - x}^{-k - 1}}} \\[5mm] & = \color{#f00}{1 \over k\,\, k!} \end{align}

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For $m\in\mathbb{N}$, define the sequence $s:\mathbb{N}\rightarrow\mathbb{R}$ via the infinite series

$$s_{m}:=\sum_{n=1}^{\infty}\frac{\left(n-1\right)!}{\left(m+n\right)!}.$$

Note that for each nonnegative integer $n\in\mathbb{N}\cup\{0\}$, we can express the factorial $n!$ in terms of the gamma function with the relation

$$n!=\Gamma{\left(n+1\right)}.$$

A related function known as the beta function may be defined via the integral representation

$$\operatorname{B}{\left(a,b\right)}:=\int_{0}^{1}t^{a-1}\left(1-t\right)^{b-1}\,\mathrm{d}t;~~~\small{a>0\land b>0}.$$

The beta function can be reduced to gamma functions through the following identity:

$$\operatorname{B}{\left(a,b\right)}=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(b\right)}}{\Gamma{\left(a+b\right)}}.$$


Now, for each positive integer $m\in\mathbb{N}$, we find using the technique of summing under the integral sign that

$$\begin{align} s_{m} &=\sum_{n=1}^{\infty}\frac{\left(n-1\right)!}{\left(m+n\right)!}\\ &=\sum_{k=0}^{\infty}\frac{k!}{\left(m+k+1\right)!}\\ &=\sum_{k=0}^{\infty}\frac{\Gamma{\left(k+1\right)}}{\Gamma{\left(m+k+2\right)}}\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\sum_{k=0}^{\infty}\frac{\Gamma{\left(k+1\right)}\,\Gamma{\left(m+1\right)}}{\Gamma{\left(m+k+2\right)}}\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\sum_{k=0}^{\infty}\operatorname{B}{\left(k+1,m+1\right)}\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\sum_{k=0}^{\infty}\int_{0}^{1}t^{k}\left(1-t\right)^{m}\,\mathrm{d}t\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\int_{0}^{1}\sum_{k=0}^{\infty}t^{k}\left(1-t\right)^{m}\,\mathrm{d}t\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\int_{0}^{1}\frac{\left(1-t\right)^{m}}{1-t}\,\mathrm{d}t\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\int_{0}^{1}x^{m-1}\,\mathrm{d}x;~~~\small{\left[1-t=x\right]}\\ &=\frac{1}{\Gamma{\left(m+1\right)}}\cdot\frac{1}{m}\\ &=\frac{1}{m!\,m}.\blacksquare\\ \end{align}$$


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This is not an answer but it is too long for a comment.

Considering the partial sums $$S_m^{(k)}=\sum_{n=1}^m \frac{(n-1)!}{(n+k)!}$$ and using the same ideas as David H's in his answer, we can find a nice and simple expression $$S_m^{(k)}=\frac{1}{k^2}\left(\frac{1}{\Gamma (k)}-k\,\frac{\Gamma (m+1)}{\Gamma (k+m+1)}\right)$$ which seems to be valid for any $k$ (integer or not).

If $m\to \infty$, there is convergence only if $k>0$ because of the second term in brackets (Stirling approximation shows this point easily). In such a case, $$\lim_{m\to \infty } \, S_m^{(k)}=\frac{1}{k^2\,\Gamma (k)}=\frac{1}{k\,\Gamma (k+1)}$$ which leads to David H's if $k$ is an integer.

For small values of $k$, the asymptotics is given by $$\lim_{m\to \infty } \, S_m^{(k)}=\frac{1}{k}+\gamma +\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) k+O\left(k^2\right)$$

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