I was reading a blog entry which suggests the following upper and lower bound for a binomial coefficient:
$$\left(\frac{n}{k}\right)^k \le {n \choose k} \le \left(\frac{en}{k}\right)^k$$
I found an excellent explanation of the proof here.
From this article, if $k < \sqrt{n}$:
$$\frac{n^k}{4(k!)} \le {n \choose k}\le \frac{n^k}{k!}$$
I found this reference to using the binary entropy function and Stirling's approximation.
Are these the best known upper and lower bounds? Are there any other well known tighter upper and lower bounds available?
Would I be correct in assuming that Stirling's Approximation leads to the tightest upper and lower bound?
Edit: Added 1 more inequality that I hadn't seen before with the source.
Best Answer
For the special case $n = 2k$, we can get narrower bounds:
$$\frac{1}{2k} 2^{2k} \leq \binom{2k}{k} \leq 2^{2k}$$
The proof is given as an exercise in Robert J. Vanderbei, Linear Programming: Foundations and Extensions, ch. 4.