Prove the limit for a series of products.

Question

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 $$\sum_{n=1}^\infty\prod_{k=1}^{n} \frac{k}{2 + k} = 1$$

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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*}

Answer 1

Here are some useful facts.

(1.) For every $n\geqslant0$ and $(x_k)_{0\leqslant k\leqslant n}$, one has $ \displaystyle\sum_{k=0}^n\left((1-x_k)\cdot\prod_{i=0}^{k-1}x_i\right)=1-\prod_{k=0}^{n}x_k. $

(2.) For every $(x_n)_{n\geqslant0}$ in $(0,1)$, one has $ \displaystyle\sum_{n=0}^{+\infty}\left((1-x_n)\cdot\prod_{k=0}^{n-1}x_k\right)=1-\prod_{n=0}^{+\infty}x_n. $

(3.) For every $b\gt a\gt0$, one has $ \displaystyle\prod_{n=0}^{+\infty}\left(1-\frac{a}{b+n}\right)=0. $

Apply (2.) to $x_n=\dfrac{n+1+\beta}{\lambda+n+\beta}$ for every $n\geqslant0$. Then, $$ (1-x_n)\cdot\prod_{k=0}^{n-1}x_k=\frac{\lambda-1}\beta\cdot\prod_{k=0}^{n} \frac{k+\beta}{\lambda + k + \beta}, $$ and the infinite product on the RHS of (2.) is $$ \prod_{n=0}^{+\infty}x_n=\prod_{n=0}^{+\infty}\frac{n+1+\beta}{\lambda+n+\beta}=\prod_{n=0}^{+\infty}\left(1-\frac{\lambda-1}{\lambda+n+\beta}\right)=0, $$ since $\lambda-1\gt0$, using (3.). Hence $\dfrac{\lambda-1}\beta$ times the LHS of the identity in the post is $1-0=1$, as desired.

Edit: To compute $\displaystyle\sum_{n=1}^\infty\prod_{k=1}^{n} \frac{k}{2 + k} = 1$, one can note more simply that $$ \prod_{k=1}^{n} \frac{k}{2 + k}=\frac{2}{(n+1)(n+2)}=\frac{2}{n+1}-\frac{2}{n+2}. $$ Edit: To see that (3.) holds, note that $1-x\leqslant\mathrm e^{-x}$ for every real number $x$, hence $$ \prod_{n=0}^{+\infty}\left(1-\frac{a}{b+n}\right)\leqslant\exp\left(-\sum_{n=0}^{+\infty}\frac{a}{b+n}\right)=0. $$

Answer 2

First, let's agree that $$ \prod_{k=0}^n \frac{k+\beta}{k+\lambda+\beta} = \frac{(\beta)_{n+1}}{(\beta+\lambda)_{n+1}} = \frac{\Gamma(\beta+n+1) \Gamma(\beta+\lambda)}{\Gamma(\beta)\Gamma(\beta+\lambda+n+1)} = \frac{B(\lambda, \beta+n+1)}{B(\lambda, \beta)} $$ where $(a)_n$ denotes a Pochhammer symbol, and $B(a,b)$ denotes Euler's beta function. Using Euler integral representation: $$ \prod_{k=0}^n \frac{k+\beta}{k+\lambda+\beta} = \frac{1}{B(\lambda,\beta)} \int_0^1 u^{\beta+n} (1-u)^{\lambda-1} \mathrm{d} u $$ Interchanging the summation and integration: $$ \sum_{n=0}^\infty \prod_{k=0}^n \frac{k+\beta}{k+\lambda+\beta} = \frac{1}{B(\beta,\lambda)} \int_0^1 (1-u)^{\lambda-2} u^{\beta} \mathrm{d} u = \frac{B(\beta+1,\lambda-1)}{B(\beta,\lambda)} = \frac{\beta}{\lambda-1} $$

Answer 3

Your identity can be expressed as a Gaussian hypergeometric function of unit argument:

$$\sum_{n=0}^\infty \prod_{k=0}^n \frac{\beta+k}{\beta+\lambda+k}=\frac{\beta}{\beta+\lambda}\sum_{n=0}^\infty \frac{(\beta+1)_n (1)_n}{(\beta+\lambda+1)_n}\frac{1^n}{n!}=\frac{\beta}{\beta+\lambda}{}_2 F_1\left({{\beta+1,1}\atop{\beta+\lambda+1}}\mid 1\right)$$

Now, there is Gauss's hypergeometric identity (see this as well):

$${}_2 F_1\left({{a,b}\atop{c}}\mid 1\right)=\frac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)},\qquad \Re(c-a-b)>0$$

which, when applied to your special case, gives

$$\require{cancel}\begin{align*} \frac{\beta}{\beta+\lambda}{}_2 F_1\left({{\beta+1,1}\atop{\beta+\lambda+1}}\mid 1\right)&=\frac{\beta}{\beta+\lambda}\frac{\Gamma(\beta+\lambda+1-1-(\beta+1))\Gamma(\beta+\lambda+1)}{\Gamma(\beta+\lambda+1-1)\Gamma(\beta+\lambda+1-(\beta+1))}\\ &=\frac{\beta}{\beta+\lambda}\frac{\Gamma(\lambda-1)\Gamma(\beta+\lambda+1)}{\Gamma(\beta+\lambda)\Gamma(\lambda)}\\ &=\frac{\beta}{\lambda-1}\frac{\cancel{\Gamma(\lambda-1)}\cancel{\Gamma(\beta+\lambda+1)}}{\cancel{\Gamma(\beta+\lambda+1)}\cancel{\Gamma(\lambda-1)}}=\frac{\beta}{\lambda-1} \end{align*}$$

Answer 4

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.