Infinite series of a product

Question

Let $a_n = \prod_{i=0}^{n-1} (r+i)$, with $0<r<1$. I want to find the infinite series $$ \sum_{n=0}^{\infty} n\,\frac{a_n}{n!}. $$

I tried to look at the bounds $ r^n < a_n < (r+n-1)^n$, but the upper bound is not a good one, because $\sum_{0}^{\infty} n\,\frac{(r+n-1)^n}{n!}$ diverges. The ratio test also doesn't work because $$\lim_{n\rightarrow \infty} \frac{a_{n+1}/(n+1)!}{a_n/n!}=1.$$ Is the above series convergent? If so, what is the value of that?

Answer 1

The product can be written as, $$ \displaystyle \prod_{i=0}^{n-1}(r+i)=r(n-1)!\binom{r+n-1}{n-1} $$ Therefore the sum is, $$ \displaystyle S = r\sum_{n=0}^\infty \binom{r+n}{n}$$ The partial sum gives us, $$ \displaystyle S_k =r\sum_{n=0}^k \binom{r+n}{n} = r\binom{k+r+1}{k} $$ So it seems it would diverge very quickly.

Answer 2

By comparing individual terms, we easily have, for $n\ge 1$ $$a_n > r\cdot 1\cdot 2\cdots (n-1) = r(n-1)!.$$

Thus each term in the sum (after $n=0$) is greater than $r>0$, so the series diverges by comparison to $\sum_{n=1}^\infty 1$.

Answer 3

Using Pochhammer symbols $$a_n = \prod_{i=0}^{n-1} (r+i)=r (r+1)_{n-1}$$ $$b_n=n \frac {a_n}{n!}=\frac{\Gamma (n+r)}{\Gamma (n) \,\Gamma (r)}\implies \frac{b_{n+1}}{b_n}=1+\frac r n \quad >0 \quad \forall r,n$$ $$S_p=\sum_{n=0}^p b_n=\frac{p \,\Gamma (p+r+1)}{(r+1)\, \Gamma (p+1)\,\Gamma (r)}$$ Using Stirling approximation $$S_p=\frac{p^{1+r}}{\Gamma (r) } \Bigg[\frac{1}{(r+1) }+\frac{r}{2 p}+ O\left(\frac{1}{p^2}\right)\Bigg]$$