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The numbers $\binom{n+\alpha-1}{n}$, $n\geq0$; $\alpha\in\mathbb{R}$, arise as the coefficients of the Taylor series of the holomorphic function $(1-z)^{-\alpha}$ around $z=0$. When $\alpha\not\in\{0,-1,-2,\ldots\}$ we can appeal to the Gamma function $\Gamma$ to show that $$ \binom{n+\alpha-1}{n}\sim \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{n^{1-\alpha}}\,,\ \text{ when }n\to\infty, $$ where, for $x\in\mathbb{R}$ and $k\geq0$ integer,$$\binom{x}{k}:=\frac{x(x-1)(x-2)\cdots(x-k+1)}{k!}.$$

Using the Gamma function seems kind of an overkill to me. Are there a more elementary way to obtain this, maybe just up to the constant $1/\Gamma(\alpha)$?

Any contribution is appreciated, thank you!!

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    How is $\binom{n+\alpha-1}{n}$ defined for $\alpha\in\mathbb{R}$ without the $\Gamma$ function? – 2017-02-05
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    @JackD'Aurizio Sorry, I could not reply sooner. It is indeed like in your answer, I am editing the question to avoid future confusions. – 2017-02-05

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Avoiding the $\Gamma$ function: $$\binom{n+\alpha-1}{n} = \frac{1}{n!}\prod_{k=0}^{n-1}(n+\alpha-1-k) = \prod_{k=1}^{n}\left(1+\frac{\alpha-1}{k}\right) $$ and if we approximate $\left(1+\frac{\alpha-1}{k}\right)$ with $\exp\left(\frac{\alpha-1}{k}\right)$ the last product turns into $$\exp\left((\alpha-1) H_n\right)\approx e^{\gamma}\cdot n^{\alpha-1}. $$

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    Amazing answer, thank you!! Just one concern; how can I prove that the last approximation is indeed correct? That is, we are approximating each term in an infinite product, so how can we guarantee that the product of the approximated terms actually approximates the initial product? – 2017-02-05
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    @mathbeing: you may be a little more accurate and use two terms of the Taylor series of $\log(1+x)=x-\frac{x^2}{2}$ to check the previous approximation is off by at most a constant factor. – 2017-02-05
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Binomial coefficients:

$$\binom{n+\alpha-1}n=\frac{(n+\alpha-1)!}{n!(\alpha-1)!}=\frac1{(\alpha-1)!}(n+1)(n+2)\dots(n+\alpha-2)(n+\alpha-1)\\\sim\frac1{(\alpha-1)!}n^{\alpha-1}$$

if $\alpha\in\mathbb R$,

$$\binom{n+\alpha-1}n=\frac{(\alpha)(\alpha+1)\dots(\alpha+n-1)}{n!}\sim\frac1{n!}\alpha^{n-1}$$

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    What is $(n+1)(n+2)\cdots(n+\alpha-1)$ supposed to mean when $\alpha$ is, for instance, $1/2$? – 2017-02-05
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    @mathbeing sorry for the late response, I've updated my answer. :-) – 2017-02-05
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    I might be experiencing some "lack of coffee", but I just don't see why the last estimation holds or how does it show that the asymptotic grow order of $\binom{n+\alpha-1}{n}$ is $n^{\alpha-1}$. – 2017-02-05
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    @mathbeing does it make sense to say $\alpha(\alpha+1)\sim\alpha^2$? How about $\alpha(\alpha+1)(\alpha+2)\sim\alpha^3$? – 2017-02-05
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    Of course, but then it should be $\alpha^n$ as there are $n$ products... And how do you derive the order $n^{\alpha-1}$ from there? – 2017-02-05
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    The first one has $\alpha-1$ products containing $n$ while the second one has $n-1$ products containing $\alpha$. Do you see that? – 2017-02-05
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    I have no problem with the case when $\alpha$ is integer. My concern is about the very last assertion of your answer ("$\sim\alpha^{n-1}/n!$") and how does it imply the estimate in my question. There are not $n-1$ products containing $\alpha$, but $n$. However, even if you meant "$\sim\alpha^{n}/n!$", I don't see how that would imply the estimate in my question. (Sorry I didn't reply any sooner, it is a headache to write in my phone). – 2017-02-05
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    @mathbeing Oh, you are specifically interested in the first equivalence... hm... I'd look towards the Gamma function by natural instinct, and Jack's answer seems the best alternative. – 2017-02-06