upper bound formula the binomial coefficients with real valued arguments I'm trying to prove the following.
Let $n\in\mathbb{N},m\in\mathbb{N}\cup\{0\},\alpha\in (n-1,n)$ and $N\in\mathbb{N}:N\ge m+1$.
Prove that
\begin{align} 
&\sum_{k=N+1}^\infty\Big{|}\binom{n+m-\alpha}{k}(-1)^k\Big{|}\le\sum_{k=N+1}^\infty \frac{\exp{((n+m-\alpha)^2+n+m-\alpha})}{k^{n+m+1-\alpha}}\le\\
&\le\int_N^\infty \frac{\exp{((n+m-\alpha)^2+n+m-\alpha})}{k^{n+m+1-\alpha}}dk=\frac{\exp{((n+m-\alpha)^2+n+m-\alpha})}{N^{n+m-\alpha}(n+m-\alpha)}
\end{align}
The first $\le$ is tricky for me.
I have found out for the second $\le$ so called the integral test, but not sure is it the right approach.
 A: As mentioned in the comments, to show the first inequality we need to show that $\binom{n}{k}\leq 2^{n^2+n}/k^{n+1}$. Now observe 
\begin{align*}
\log \Big(\frac{n^n}{k^k (n-k)^{n-k}} \Big)&=n\log n-k\log k -(n-k)\log (n-k) \\
&=(n-k)\log n -(n-k)\log (n-k) +k\log n-k\log k\\
&=-(n-k)\log (\frac{n-k}{n}) -k\log (\frac{k}{n})%=nH(k/n),
\end{align*}
where the second equality follows by adding and subtracting the term $k\log n$ and the last equality follows by definition of the Binary entropy function.
 Finally we make the observation that
$$
  \binom{n}{k}=\frac{n!}{k! (n-k)!}
\approx \frac{\sqrt{n}}{2\pi \sqrt{k(n-k)}} \frac{n^n}{k^k (n-k)^{n-k}} 
  =\frac{\sqrt{n}}{\sqrt{2\pi k(n-k)}}2^{nH(k/n)}.
$$
using Stirling approximation.
Hence it remains to show that $\log\binom{n}{k}\approx nH(k/n)\leq n^2-n\log k$. Indeed this is true since $H(k/n)\leq 1\leq n-\log n\leq n-\log k$, which proves the inequality for $n>3$ and using $H(\alpha)< 1$. ( in fact we obtain a stronger upper bound that desired).
