How to use Euler-Maclaurin summation to show the following relationship over Gamma function I encountered this problem that I could not figure out, the first form of the Euler Maclaurin summation is:
$$
\sum_{a \leq k \leq b}f(k) = \int_a^b f(x)dx + \frac{f(a)+f(b)}{2} + \sum_{1 \leq i \leq m}\frac{B_{(2i)}}{(2i)!}f^{(2i-1)}(x)|^b_a
$$
while its second form is:
$$
\sum_1^N f(k) = \int_1^Nf(x)dx + \frac{f(N)}{2} + \gamma + \sum_{1\leq k \leq m} \frac{B_{2i}}{(2i)!}f^{(2i-1)}(N)+R_m
$$
where for both forms $B_i$ represents the $i$th Bernoulli number and $f^{(i)}$ represents the $i$th derivative.
Using (either of ) these formulas, am supposed so show that given the Gamma function $\Gamma(n)=(n-1)!$, we have the following relationship over generalized binomial coefficients:
$$
\binom{n+\alpha}{n} \equiv \frac{\Gamma(n+\alpha+1)}{\Gamma(n+1)\Gamma(\alpha+1)} \sim \frac{n^\alpha}{\Gamma(\alpha+1)}
$$
If I use the definition $\Gamma(n)=(n-1)!$, the first equivalence $\equiv$ relation could be derived easily ... But I could not figure out how the Euler Maclaurin equation comes to play ... I am having a hunch it has something to do with proving the $\sim$ relation... Any help ?
 A: I think you can get this from the Bernoulli integral for the $\Gamma$ function:
$$
\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx.
$$
If we use the first Euler Maclaurin summation and swap the integral and summation, we get:
$$
\int_a^b f(x) dx = \sum_{a\leq k\leq b} f(k) - \frac{f(a) + f(b)}{2} - \sum_{1\leq i\leq m} \frac{B_{2i}}{(2i)!} f^{(2i-1)}(N) - R(m).
$$
We then truncate the last expansion term summation leaving the simple (0-th order "expansion"):
$$
\int_a^b f(x) dx \approx \sum_{a\leq k\leq b} f(k) - \frac{f(a) + f(b)}{2}.
$$
For our $\Gamma$ function integral, we let $f(x)=x^{z-1}e^{-x}$ to get:
$$
\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx
\approx \sum_{0\leq k\leq \infty} k^{z-1} e^{-k} - \lim_{a\downarrow 0}\frac{a^{z-1}}{2e^a} - \lim_{b\to\infty} \frac{b^{z-1}}{2e^b} = \sum_{0\leq k\leq \infty} k^{z-1} e^{-k}.
$$
Note that for the above to hold, we require $z>0$ so that $f(x)$ is bounded above or $z$ such that $f$ is dominated by a function $g$ with a finite integral $\int_0^\infty g(x) dx$.
Now rewrite $\binom{n+\alpha}{n}$:
$$
\binom{n+\alpha}{n} = \frac{\Gamma(n+\alpha+1)}{\Gamma(n+1)\Gamma(\alpha+1)}
\approx \frac{\sum_{0\leq k\leq \infty} k^{n+\alpha} e^{-k}}{\Gamma(\alpha+1)\sum_{0\leq k\leq \infty} k^{n} e^{-k}}
= \frac{k^\alpha}{\Gamma(\alpha+1)} \frac{\sum_{0\leq k\leq \infty} k^{n} e^{-k}}{\sum_{0\leq k\leq \infty} k^{n} e^{-k}}
= \frac{k^\alpha}{\Gamma(\alpha+1)}.
$$
