Closed-form value of a product $\prod_{i = 2}^n (1 + \frac{p}{i})$ with $0 \leq p \leq 2$

Question

Let $f(n+1;p) = \prod_{i = 2}^n (1 + \frac{p}{i})$, where $0 \leq p \leq 2$ and $n \geq 2$ with $f(2;p)=1,\forall p$.

We have $f(n;0) = 1$, $f(n; 1)=n/2$. I also can see that $f(n; 2) \leq n^2/4$.

Is it possible to have a closed-form value of $f(n;p)$ for general $p$ or at least for $p = 2$?

Answer 1

We can write $$f(n;p)=\frac{\prod_{i=2}^{n-1}(i+p)}{\prod_{i=2}^{n-1}i}=\frac{\prod_{i=2}^{n-1}(i+p)}{(n-1)!}.$$ Using the property of the Gamma function that $\Gamma(x+1)=x\Gamma(x)$ for every $x$, we have $$\prod_{i=2}^{n-1}(i+p)=\prod_{i=2}^{n-1}\frac{\Gamma(i+1+p)}{\Gamma(i+p)}=\frac{\Gamma(n+p)}{\Gamma(2+p)},$$ so $$f(n;p)=\frac{\Gamma(n+p)}{\Gamma(2+p)\Gamma(n)}=\frac1{p(1+p)\mathrm B(n,p)},$$ where $\mathrm B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is the Beta function.

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If $p$ is an integer, this gives an expression in terms of factorials: $$f(n;p)=\frac{(n+p-1)!}{(p+1)!(n-1)!}=\frac1{p+1}\binom{n+p-1}{p}.$$ In particular, $$f(n;2)=\frac13\binom{n+1}2=\frac{n(n+1)}{6}.$$