# bound for binomial coefficients

How can one show $\displaystyle\frac{(m+n-1)!}{m ! (n-1)!}\leq \left[\frac{e (m+n-1)}{n}\right]^{n-1}$ ?

• Have you tried Stirling's approximation? – S. Carnahan Jun 12 '10 at 7:50
• Surely this is homework question. I did a course this year and that was the first question on the first problem sheet. – alext87 Jun 12 '10 at 7:55
• I did try Stirling's approximation but am unable to get this expression. – Vagabond Jun 12 '10 at 8:29
• The first and obvious approach is to use induction on $n$. In dx.doi.org/10.1007/s11139-006-0075-1 (see also arXiv.org/abs/math/0304021) I had a similar estimate; there however the deal was bout the beta-integral. So, if you take the reciprocal of both sides you can use the estimate from that paper (it's quite sharp). – Wadim Zudilin Jun 12 '10 at 8:38

Denote the quotient of the right and left hand sides, $$f(m,n)=\biggl(\frac{e(m+n-1)}n\biggr)^{n-1}\bigg/\binom{m+n-1}m.$$ Then $f(m,1)=1$ for all $m\in\mathbb N$ and $$\frac{f(m,n+1)}{f(m,n)} =\frac{e}{\biggl(1+\dfrac1n\biggr)^n}\cdot\biggl(1+\frac1{m+n-1}\biggr)^{n-1} > 1,$$ that is, $$f(m,n+1)> f(m,n)>\dots> f(m,2)> f(m,1)=1.$$ This proves the required inequality.
Another immediate proof can be obtained from $$\frac{(m+n-1)!}{m!}\le(m+n-1)^{n-1}$$ (which is obvious) and $$\left(\frac{n}{e}\right)^{n-1}\le(n-1)!$$ which after multiplying by $n$ and taking logs becomes $$n\log(n)-n+1\le\sum_{k=2}^{n}\log(k)$$ which is immediate as the RHS is an obvious upper bound for $$\int_1^n\log(x)\,dx=(x\log(x)-x)|_1^n=n\log(n)-n+1.$$
• Vladimir, I have a special smile for this: $\dddot\smile$ – Wadim Zudilin Jun 14 '10 at 12:02