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I have trouble in getting a closed form expression of $\binom{p+n}{n}$, otherwise written as

$$\Big(1+\frac{p}{1}\Big)\Big(1+\frac{p}{2}\Big) ... \Big(1+\frac{p}{n}\Big)$$

where $p \in [0,1]$. Even an approximation would suffice for me.

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    Where does this come from? Add any context you have and your efforts as well. – 2017-02-02

2 Answers 2

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My two cents would be to multiply it out. The sum is approximately

$$1 + \frac{p(\frac{1}{1}+\frac{1}{2} ...+ \frac{1}{n})}{1!} + \frac{p^2(\frac{1}{1}+\frac{1}{2} ...+ \frac{1}{n})^2}{2!}+...\frac{p^n(\frac{1}{1}+\frac{1}{2} ...+ \frac{1}{n})^n}{n!}$$

which is approximately equal to $$e^{p(\frac{1}{1}+\frac{1}{2} ...+ \frac{1}{n})}$$ as $n$ becomes large. If we take the limit to infinity it's $e^{p(\log n +\gamma)}= (e^{\gamma} n)^p$.

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Write it as $$\frac{(1+p)(2+p)\ldots(n+p)}{n!} = \frac{\Gamma(n+p+1)}{n! \;\Gamma(p+1)} $$

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    Sorry, I don't see how this is helpful. Could you elaborate more in your answer? Thanks. – 2017-02-02
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    @S.C.B. It's a closed form for what learner was looking for, isn't it? Multiplying the second form learner wrote in his post by $\frac{n!}{n!}$ gives the expression on the LHS in RobertIsrael's post. The RHS is as closed form as you'll get since $p \in [0,1]$ unless $p=0$ or $p=1$. – 2017-02-02
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    @CodeLabMaster That depends on how you define closed forms. I think that depending, you could say that $$\binom{n}{k}$$ it self is a closed form? – 2017-02-02
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    @S.C.B. I agree that it's a closed form, but if it was not acknowledged as one in the OP, then this seems like the next step, writing it with Gamma. – 2017-02-02
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    Thanks so much Robert. I was looking into an exponential form though so I can get the approximate value of p. – 2017-02-02