I need help to evaluate $\sum_{k=1}^{\infty}\frac{1}{\prod_{j=0}^p(k+j)}$

Question

I need help to evaluate the following sum: $$\sum_{k=1}^{\infty}\frac{1}{\prod_{j=0}^p(k+j)}.$$

I wrote it in terms of Gamma functions to eliminate the product: $\sum_{k=1}^{\infty}\frac{\Gamma(k)}{\Gamma(p+k+1)}$. I'm not sure what to do from here.

Answer 1

Since the fraction looks almost like Beta function it was worth to complete it. So we have $$\begin{split} \sum_{k=1}^{\infty}\frac{\Gamma(k)}{\Gamma(p+k+1)} & = \frac{1}{\Gamma(p+1)}\times \sum_{k=1}^{\infty}\frac{\Gamma(k)\Gamma(p+1)}{\Gamma(p+k+1)} \\ & = \frac{1}{\Gamma(p+1)}\times \sum_{k=1}^{\infty}\mathrm{B}(p+1,k)\\ & = \frac{1}{\Gamma(p+1)}\times \sum_{k=1}^{\infty}\int_{0}^{1} \xi^{k-1}(1-\xi)^{p}\,\mathrm{d}\xi\\ & = \frac{1}{\Gamma(p+1)}\times \int_{0}^{1} \sum_{k=1}^{\infty}\xi^{k-1}(1-\xi)^{p}\,\mathrm{d}\xi\\ & = \frac{1}{\Gamma(p+1)}\times \int_{0}^{1} (1-\xi)^{p-1}\,\mathrm{d}\xi\\ &= \frac{1}{\Gamma(p+1)}\times \frac{1}{p} \ . \end{split}$$ If you do not like Beta function you might try do this by partial-ish fraction decomposition and telescoping since $$\frac{1}{k(k+1)\cdot\ldots\cdot(k+p)}=\frac{1}{p}\left(\eta_{k,p}-\eta_{k+1,p}\right),$$ where $\eta_{k,p}=\frac{1}{k(k+1)\cdot\ldots\cdot (k+p-1)}$.

Answer 2

It is well known that both $\dfrac1{\prod_{j=0}^n(k+j)} $ and $\prod_{j=0}^n(k+j)$ have a decomposition that results in their sums telescoping.

Let $p_n(k) =\prod_{j=0}^n(k+j) $. Then

$\begin{array}\\ p_n(k+1)-p_n(k) &=\prod_{j=0}^n(k+1+j)-\prod_{j=0}^n(k+j)\\ &=\prod_{j=1}^{n+1}(k+j)-\prod_{j=0}^n(k+j)\\ &=\prod_{j=1}^{n}(k+j)((n+k+1)-k)\\ &=\prod_{j=0}^{n-1}(k+1+j)(n+1)\\ &=(n+1)p_{n-1}(k+1)\\ \end{array} $

so $p_n(k) =\dfrac{p_{n+1}(k)-p_{n+1}(k-1)}{n+2} $.

Similarly,

$\begin{array}\\ \dfrac1{p_n(k)}-\dfrac1{p_n(k+1)} &=\dfrac1{\prod_{j=0}^n(k+j)}-\dfrac1{\prod_{j=0}^n(k+1+j)}\\ &=\dfrac1{\prod_{j=0}^n(k+j)}-\dfrac1{\prod_{j=1}^{n+1}(k+j)}\\ &=\dfrac1{\prod_{j=0}^{n+1}(k+j)}((n+1+k)-k)\\ &=p_{n+1}(k)(n+1)\\ \end{array} $

so $p_n(k) =\dfrac1{n}(\dfrac1{p_{n-1}(k)}-\dfrac1{p_{n-1}(k+1)}) $.

Answer 3

What you could do is to consider the partial sum$$S_n=\sum_{k=1}^{n}\frac{\Gamma(k)}{\Gamma(p+k+1)}$$ and prove that $$S_n=\frac{(p+1) \Gamma (n+p+2)-(n+p+1) \Gamma (n+1) \Gamma (p+2)}{p \Gamma (p+2) \Gamma (n+p+2)}$$ which, once simplified is just $$S_n=\frac{1}{p \,\Gamma (p+1)}-\frac{\Gamma (n+1)}{p\, \Gamma (n+p+1)}$$ Now, using Stirling approximation twice and continuing with Taylor series for large values of $n$

$$A=\log\Bigg[\frac{\Gamma (n+1)}{p\, \Gamma (n+p+1)}\Bigg]=-(p \log (n)+\log (p))-\frac{p (p+1)}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$\frac{\Gamma (n+1)}{p\, \Gamma (n+p+1)}=e^A=n^{-p} \left(\frac{1}{p}-\frac{p+1}{2 n}+O\left(\frac{1}{n^2}\right)\right)$$ So, as a first approximation $$S_n=\frac{1}{p \,\Gamma (p+1)}-n^{-p} \left(\frac{1}{p}-\frac{p+1}{2 n}+O\left(\frac{1}{n^2}\right)\right)$$