TITLE: Prove the limit for a series of products.
QUESTION [6 upvotes]: Let $\beta > 0$, $\lambda > 1$. Show the identity
$$\sum_{n=0}^\infty\prod_{k=0}^{n} \frac{k+\beta}{\lambda + k + \beta} = \frac{\beta}{\lambda - 1}$$
I have checked the statement numerically.
The special case $\beta = 1$, $\lambda = 2$ looks like this
$$\sum_{n=1}^\infty\prod_{k=1}^{n} \frac{k}{2 + k} = 1$$

This series arises in a probabilistic setting. Let $(Y_k)_{i\ge0}$ be independent exponentially distributed variables with parameters $k+\beta$ respectively and set $S_n := \sum_{k=0}^n Y_i$. For $t \ge 0$ let $X(t) := \#\{n \ge 0: S_n < t\}$. Let $Z_\lambda$ be exponentially distributed with parameter $\lambda$ and independent of $X(t)$. Then
\begin{align*}
EX(Z_\lambda) &= E\#\{n \ge 0: S_n < Z_\lambda\} \\
& = \sum_{n=0}^\infty P(S_n < Z_\lambda) \\
& = \sum_{n=0}^\infty Ee^{-\lambda S_n} \\
& = \sum_{n=0}^\infty \prod_{k=0}^n Ee^{-\lambda Y_k} \\
& = \sum_{n=0}^\infty \prod_{k=0}^n \frac{k+\beta}{\lambda + k + \beta}
\end{align*}

REPLY [2 votes]: We want to evaluate :
$$f(\beta,\lambda)=\frac {\beta}{\lambda+\beta}+\frac {\beta(\beta+1)}{(\lambda+\beta)(\lambda+\beta+1)}+\frac {\beta(\beta+1)(\beta+2)}{(\lambda+\beta)(\lambda+\beta+1)(\lambda+\beta+2)}+\cdots$$
Setting $a:=\lambda+\beta$ this becomes :
$$F(\beta,a)=\frac {\beta}{a}+\frac {\beta(\beta+1)}{a(a+1)}+\frac {\beta(\beta+1)(\beta+2)}{a(a+1)(a+2)}+\cdots$$
This may be represented as a $\;_2F_1$ hypergeometric series since :
$$_2F_1(\beta,1;a;1)=1+\frac {\beta\cdot 1}{a\cdot 1!}+\frac {\beta(\beta+1)\cdot 1\cdot 2}{a(a+1)\cdot 2!}+\frac {\beta(\beta+1)(\beta+2)\cdot 1\cdot 2\cdot 3}{a(a+1)(a+2)\cdot 3!}+\cdots$$
but this is simply (applying Gauss' theorem) :
$$_2F_1(a,b;c;1)=\frac {\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$
so that :
$$_2F_1(\beta,1;a;1)=\frac {\Gamma(a)\Gamma(a-\beta-1)}{\Gamma(a-\beta)\Gamma(a-1)}=\frac {a-1}{a-\beta-1}=\frac {\lambda+\beta-1}{\lambda-1}$$
minus $1$ gives $\frac {\beta}{\lambda-1}$ as wished.