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*}